### Binomial Theorem

- The expansion of (a + b)
^{n}has n + 1 terms. - In the kth term of (a + b)
^{n}, the exponent of b is k – 1 and the exponent of is n – (k – 1). - In the kth term, the coefficient has k – 1 factors in the numerator and the denominator.
- You can always find the coefficients of the expansion or of a particular term by writing out Pascal’s triangle and using row n.
- The coefficients of the expansion are symmetrical around the middle term. So once you have found the first half of the coefficients, you already have the coefficients for the rest of the terms. They are the same ones you have found but listed in reverse order.

### Binomial Theorem

^{5}

- You can expand this problem using several properties of the binomial theorem.
- Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
- (x + y)
^{5}= ___(xy) + ___(xy) + ___(xy) + ___(xy) + ____(xy) + ____(xy) - Step 2 - Distribute the power n, notice how you start at 5 with x, and you start at 0 with y
- The sum of the powers of x and y always equal n, in this case 5
- (x + y)
^{5}= ___(x^{}y^{}) + ___(x^{}y^{}) + ___(x^{}y^{}) + ___(x^{}y^{}) + ____(x^{}y^{}) + ____(x^{}y^{}) - (x + y)
^{5}= ___(x^{5}y^{0}) + ___(x^{4}y^{1}) + ___(x^{3}y^{2}) + ___(x^{2}y^{3}) + ____(x^{1}y^{4}) + ____(x^{0}y^{5}) - Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
- (x + y)
^{5}= 1(x^{5}y^{0}) + 5(x^{4}y^{1}) + 10(x^{3}y^{2}) + 10(x^{2}y^{3}) + 5(x^{1}y^{4}) + 1(x^{0}y^{5}) - Step 4 : Simplify. Eliminate any powers of zero.

^{5}= x

^{5}+ 5x

^{4}y + 10x

^{3}y

^{2}+ 10x

^{2}y

^{3}+ 5xy

^{4}+ y

^{5}

^{5}

- You can expand this problem using several properties of the binomial theorem.
- Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
- (2 + y)
^{5}= ___(2y) + ___(2y) + ___(2y) + ___(2y) + ____(2y) + ____(2y) - Step 2 - Distribute the power n, notice how you start at 5 with 2, and you start at 0 with y
- The sum of the powers of 2 and y always equal n, in this case 5
- (2 + y)
^{5}= ___(2^{}y^{}) + ___(2^{}y^{}) + ___(2^{}y^{}) + ___(2^{}y^{}) + ____(2^{}y^{}) + ____(2^{}y^{}) - (2 + y)
^{5}= ___(2^{5}y^{0}) + ___(2^{4}y^{1}) + ___(2^{3}y^{2}) + ___(2^{2}y^{3}) + ____(2^{1}y^{4}) + ____(2^{0}y^{5}) - Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
- (2 + y)
^{5}= 1(2^{5}y^{0}) + 5(2^{4}y^{1}) + 10(2^{3}y^{2}) + 10(2^{2}y^{3}) + 5(2^{1}y^{4}) + 1(2^{0}y^{5}) - Step 4 : Simplify the powers of 2
- (2 + y)
^{5}= 1(32y^{0}) + 5(16y^{1}) + 10(8y^{2}) + 10(4y^{3}) + 5(2y^{4}) + 1(1y^{5}) - Step 5 - Simplify completely
- (2 + y)
^{5}= 32 + 80y + 80y^{2}+ 40y^{3}+ 10y^{4}+ y^{5} - or (2 + y)
^{5}= y^{5}+ 10y^{4}+ 40y^{3}+ 80y^{2}+ 80y + 32

^{5}= y

^{5}+ 10y

^{4}+ 40y

^{3}+ 80y

^{2}+ 80y + 32

^{5}

- You can expand this problem using several properties of the binomial theorem.
- Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
- (2x + 2y)
^{5}= ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ____(( 2x )( 2y )) + ____(( 2x )( 2y )) - Step 2 - Distribute the power n, notice how you start at 5 with (2x), and you start at 0 with (2y)
- The sum of the powers of (2x) and (2y) always equal n, in this case 5
- (2x + 2y)
^{5}= ___(( 2x )^{}( 2y )^{}) + ___(( 2x )^{}( 2y )^{}) + ___(( 2x )^{}( 2y )^{}) + ___(( 2x )^{}( 2y )^{}) + ____(( 2x )^{}( 2y )^{}) + ____(( 2x )^{}( 2y )^{}) - (2x + 2y)
^{5}= ___(( 2x )^{5}( 2y )^{0}) + ___(( 2x )^{4}( 2y )^{1}) + ___(( 2x )^{3}( 2y )^{2}) + ___(( 2x )^{2}( 2y )^{3}) + ____(( 2x )^{1}( 2y )^{4}) + ____(( 2x )^{0}( 2y )^{5}) - Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
- (2x + 2y)
^{5}= 1(( 2x )^{5}( 2y )^{0}) + 5(( 2x )^{4}( 2y )^{1}) + 10(( 2x )^{3}( 2y )^{2}) + 10(( 2x )^{2}( 2y )^{3}) + _5(( 2x )^{1}( 2y )^{4}) + 1(( 2x )^{0}( 2y )^{5}) - Step 4 : Distribute the powers using properties of exponents.
- (2x + 2y)
^{5}= 1(( 2^{5}x^{5})( 2^{0}y^{0})) + 5(( 2^{4}x^{4})( 2y )) + 10(( 2^{3}x^{3})( 2^{2}y^{2})) + 10(( 2^{2}x^{2})( 2^{3}y^{3})) + _5(( 2x )( 2^{4}y^{4})) + 1(( 2^{0}x^{0})( 2^{5}y^{5})) - Step 5 - Simplify the powers of 2
- (2x + 2y)
^{5}= 1(( 32x^{5})( 1y^{0})) + 5(( 16x^{4})( 2y )) + 10(( 8x^{3})( 4y^{2})) + 10(( 4x^{2})( 8y^{3})) + _5(( 2x )( 16y^{4})) + 1(( 1x^{0})( 32y^{5})) - Step 6 - Multiply the coefficients of Pascal's Triangle, with the coefficients of x and y.

^{5}= 32x

^{5}+ 160x

^{4}y + 320x

^{3}y

^{2}+ 320x

^{2}y

^{3}+ 160xy

^{4}+ 32y

^{5}

^{7}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (5x + 2y)
^{7}= ∑_{k = 0}^{7}(

)(5x)7 k ^{7 − k}(2y)^{k}; where (

) = [7!/(k!(7 − k)!)]7k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term=k+1 Term# = k + 1
- 6 = k + 1
- k = 5
- Use k = 5, to find the 6th term
- (

)(5x)7 5 ^{7 − 5}(2y)^{5}= (

)(5x)7 5 ^{2}(2y)^{5}= [7!/(5!(7 − 5)!)](5x)^{2}(2y)^{5}= [7*6*5!/5!(2)!](5x)^{2}(2y)^{5}= - [7*6*5!/5!(2)!](5x)
^{2}(2y)^{5}= [7*6*/(2)!](5x)^{2}(2y)^{5}= [42/2](5x)^{2}(2y)^{5}= 21(25x^{2})(32y^{5})

^{7}= 16800x

^{2}y

^{5}

^{50}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (x + y)
^{50}= ∑_{k = 0}^{50}(

)x50 k ^{50 − k}y^{k}; where (

) = [50!/(k!(50 − k)!)]50 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term=k+1
- Term# = k + 1
- 30 = k + 1
- k = 29
- Use k = 29, to find the 30th term
- (

)(x)50 29 ^{50 − 29}(y)^{29}= (

)(x)50 29 ^{21}(y)^{29}= [50!/(29!(50 − 29)!)](x)^{21}(y)^{29}= [50!/29!(21)!](x)^{21}(y)^{29}= - [50*49*48...28*29!/29!(21)!](x)
^{21}(y)^{29}= [50*49*48...28/(21)!](x)^{21}(y)^{29}= [50*49*48...28/(21)!](x)^{21}(y)^{29}=

^{50}= 67,327,446,062,800x

^{21}y

^{29}

^{7}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (6x + y)
^{7}= ∑_{k = 0}^{7}(

)( 6x )7 k ^{7 − k}y^{k}; where (

) = [7!/(k!(7 − k)!)]7 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term# = k + 1
- 4 = k + 1
- k = 3
- Use k = 3, to find the 4th term
- (

)(6x)7 3 ^{7 − 3}(y)^{3}= (

)(6x)7 3 ^{4}(y)^{3}= [7!/(3!(7 − 3)!)](6x)^{4}(y)^{3}= [7!/3!(4)!](6x)^{4}(y)^{3}= - [7*6*5*4!/3!(4)!](6x)
^{4}(y)^{3}= [7*6*5*/3!](6x)^{4}(y)^{3}= 35(6x)^{4}y^{3}= 35(6^{4}x^{4})y^{3}= 35(1296x^{4})y^{3}= 45360x^{4}y^{3}

^{7}= 45360x

^{4}y

^{3}

^{7}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (2x + ( − 4y))
^{7}= ∑_{k = 0}^{7}(

)( 2x )7 k ^{7 − k}( − 4y)^{k}; where (

) = [7!/(k!(7 − k)!)]7 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term# = k + 1
- 4 = k + 1
- k = 3
- Use k = 3, to find the 4th term
- (

)(2x)7 3 ^{7 − 3}( − 4y)^{3}= (

)(2x)7 3 ^{4}( − 4y)^{3}= [7!/(3!(7 − 3)!)](2x)^{4}( − 4y)^{3}= [7!/3!(4)!](2x)^{4}( − 4y)^{3}= - [7*6*5*4!/3!(4)!](2x)
^{4}( − 4y)^{3}= [7*6*5*/3!](2x)^{4}( − 4y)^{3}= 35(2x)^{4}( − 4y)^{3}= 35(2^{4}x^{4})( − 4)^{3}y^{3}= 35(16x^{4})( − 64)y^{3}= − 35840x^{4}y^{3}

^{7}= − 35840x

^{4}y

^{3}

^{3}− 4y

^{3})

^{7}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (2x
^{3}+ ( − 4y^{3}))^{7}= ∑_{k = 0}^{7}(

)( 2x7 k ^{3})^{7 − k}( − 4y^{3})^{k}; where (

) = [7!/(k!(7 − k)!)]7 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term=k+1
- Term# = k + 1
- 4 = k + 1
- k = 3
- Use k = 3, to find the 4th term
- (

)(2x7 3 ^{3})^{7 − 3}( − 4y)^{3}= (

)(2x)7 3 ^{4}( − 4y)^{3}= [7!/(3!(7 − 3)!)](2x^{3})^{4}( − 4y^{3})^{3}= [7!/3!(4)!](2x^{3})^{4}( − 4y^{3})^{3}= - [7*6*5*4!/3!(4)!](2x
^{3})^{4}( − 4y^{3})^{3}= [7*6*5*/3!](2x^{3})^{4}( − 4y^{3})^{3}= 35(2^{4}x^{12})( − 4^{3})(y^{9}) = 35(16x^{12})( − 64)(y^{9}) = − 35840x^{12}y^{9}

^{3}− 4y

^{3})

^{7}= − 35840x

^{12}y

^{9}

^{2}+ y

^{3})

^{14}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]n k - This particular problem would then be defined as
- (x
^{2}+ y^{3})^{14}= ∑_{k = 0}^{14}(

)( x14 k ^{2})^{14 − k}(y^{3})^{k}; where (

) = [14!/(k!(14 − k)!)]14 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term# = k + 1
- 4 = k + 1
- k = 3
- Use k = 3, to find the 4th term
- (

)(x14 3 ^{2})^{14 − 3}(y^{3})^{3}= (

)(x14 3 ^{2})^{11}(y^{3})^{3}= [14!/(3!(14 − 3)!)](x^{2})^{11}(y^{3})^{3}= [14!/3!(11)!](x^{22})(y^{9}) - [14!/3!(11)!](x
^{22})(y^{9}) = [14*13*12*/3!](x^{22})(y^{9}) = 364x^{22}y^{9}

^{2}+ y

^{3})

^{14}= 364x

^{22}y

^{9}

^{2}+ y

^{3})

^{14}

- Recall that the binomial theorem is defined as
- (a + b)
^{n}= ∑_{k = 0}^{n}(

)an k ^{n − k}b^{k}; (

) = [n!/(k!(n − k)!)]nk - This particular problem would then be defined as
- (x
^{2}+ y^{3})^{14}= ∑_{k = 0}^{14}(

)( x14 k ^{2})^{14 − k}(y^{3})^{k}; where (

) = [14!/(k!(14 − k)!)]14 k - To find any term in the binomial expansion, all you need to solve is for k.
- The relationship between Term# and k is the following
- Term=k+1
- Term# = k + 1
- 10 = k + 1
- k = 9
- Use k = 9, to find the 10th term
- (

)(x14 9 ^{2})^{14 − 9}(y^{3})^{9}= (

)(x14 9 ^{2})^{5}(y^{3})^{9}= [14!/(9!(14 − 9)!)](x^{2})^{5}(y^{3})^{9}= [14!/9!(5)!](x^{10})(y^{27}) - [14!/9!(5)!](x
^{10})(y^{27}) = [14*13*12*11*10*/(5)!](x^{22})(y^{9}) = 2002x^{10}y^{27}

^{2}+ y

^{3})

^{14}= 2002x

^{10}y

^{27}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Binomial Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Pascal's Triangle 0:06
- Expand Binomial
- Pascal's Triangle
- Properties 6:52
- Example: Properties of Binomials
- Factorials 9:11
- Product
- Example: Factorial
- Binomial Theorem 11:08
- Example: Binomial Theorem
- Finding a Specific Term 18:36
- Example: Specific Term
- Example 1: Expand 24:39
- Example 2: Fourth Term 30:26
- Example 3: Five Terms 36:13
- Example 4: Three Iterates 45:07

### Algebra 2

### Transcription: Binomial Theorem

*Welcome to Educator.com.*0000

*In today's lesson, we are going to be covering the binomial theorem.*0002

*We are going to start out by talking about what we mean when we say we are going to expand a binomial.*0007

*So, if you expand a binomial, this is what happens.*0013

*I have (a + b) ^{n}; and I am going to expand that for n = 1, 2, 3, 4.*0019

*I am going to let n equal 1; and I am going to then get (a + b) ^{1}, which is just a + b.*0027

*If I let n equal 2, then I have(a + b) ^{2}; that is going to give me, if you will recall, a^{2} + 2ab + b^{2}.*0041

*Let n equal 3: you get (a + b) ^{3}, which is a little bit more complicated to work out.*0057

*But if you figure that all out, (a + b)(a + b)(a + b), you would end up with a ^{3} + 3a^{2}b + 3ab^{2} + b^{3}.*0064

*Letting n equal 4, (a + b) ^{4} is going to give you a^{4} + 4a^{3}b +*0079

*6a ^{2}b^{2} + 4ab^{3} + b^{4}.*0091

*One more: let n equal 5: this is going to give me (a + b) ^{5}.*0099

*This equals a ^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b...*0108

*actually, this is going to be b ^{3}...+ 5ab^{4} + b^{5}.*0127

*All right, before we go on to talk about Pascal's triangle, let's note a few things about this expansion.*0136

*And these will be summed up on the next slide.*0144

*A few things that you will notice are that the first term is a ^{n}, and the last term is b^{n}.*0146

*The second thing is that the number of terms equals n + 1.*0157

*You will also notice a pattern: let's look at n = 4, (a + b) ^{4}.*0179

*The first term is a ^{n}, so it is a^{4}; in each subsequent term, the power that a is raised to (the exponent) decreases by 1.*0185

*So, I started out with a ^{4}; then here it's a^{3}; a^{2}; a; and then a is gone.*0195

*So, the exponent for a decreases by 1 for each term; b does the opposite.*0203

*There is no b here; this would be b ^{0}...it is 1; and then in my next term, I get b to the first power, b^{2}, b^{3}, b^{4}.*0213

*So, with each term, exponents for a are decreasing; exponents for b are increasing.*0223

*The next thing that you might notice is the coefficients; let's look at the coefficients.*0230

*The coefficient of the second term is equal to n; again, this is all summed up on the next slide, but this is just to introduce it here.*0234

*So, if I am looking at n = 3 (that is (a + b) ^{3}), and I expand that; then I get that the second term has a coefficient of 3.*0242

*The second term here has a coefficient of 4; the second term here has a coefficient of 5.*0252

*Pascal's triangle--we are getting to that now.*0258

*I can take the coefficient of these terms and use them to create an array that is called Pascal's triangle.*0260

*Looking at it, it actually starts out with 1; so if we let n equal 0, that is just going to give me (a + b) ^{0}, which is going to be 1.*0266

*Then, in my next row, I am going to have 1, 1, and 1; those are my coefficients.*0277

*In the next row, they are 1, 2, 1; in the next row, 1, 3, 3, 1.*0289

*Now, I can look up here to get the next row; but I don't even need to, because each number is the sum of the two numbers just above it.*0302

*So, 1 + 1 forms 2; 1 + 2 gives me 3; 2 + 1 gives me 3.*0314

*So, all I have to do, actually--these outside numbers are always 1; but I could just say 1 + 3...this is going to be 4.*0325

*3 + 3 is equal to 6; 3 + 1...that is equal to 4.*0334

*So, I look up here to verify that; and indeed, it is 1, 4, 6, 4 1.*0340

*In the next row, I would add 1, and I add 1 + 4; I am going to get 5; 4 + 6 gives me 10;*0345

*6 + 4 gives me 10; 4 + 1 gives me 5; and again, I have a 1 out here.*0352

*And that matches up with what I have for the expansion of (a + b) ^{5}.*0357

*So, Pascal's triangle is an interesting array that comes from the coefficients of the expansion of this binomial.*0363

*Something else to note is the symmetry around the middle term; you might have seen that up here, but it is even more obvious right here.*0371

*The middle term is 2; it is flanked by 1's; here I have two middle terms, because there is an even number of terms;*0377

*so the two middle terms are 3; they are flanked by 1's; now I have a middle term of 6; next to it are 4's; outside of that, 1's.*0384

*Two middle terms are 10; the next terms are 5; the next terms are 1.*0392

*And that is helpful, because if you are looking for the coefficients, you don't need to find all of them; you only need to find half of them.*0397

*And then, you can use this symmetry to know what the other half are.*0403

*Summing up these properties of the expansion of (a + b) ^{n}: there are n + 1 terms.*0409

*Let's use an example that we just talked about, (a + b) ^{5}, which gave me a^{5} +*0418

*5a ^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} +*0427

*5ab ^{4} + b^{4}, to illustrate this.*0437

*There are n + 1 terms; so here, n = 5; therefore, I am going to have n + 1 terms, equals 5 + 1, so I am going to have 6 terms.*0442

*And I do have 1, 2, 3, 4, 5, 6 terms.*0455

*The first term is going to be a ^{n} (that is a^{5}), and the last term is b^{n}, which should be b^{5} right here.*0458

*In successive terms, the exponent of a decreases by 1, while that of b increases by 1.*0470

*I start out with a ^{5}, then a^{4}, to the third, to the second, and then a.*0475

*b goes the opposite way: I start with b, b ^{2}, b^{3}, b^{4}, and b^{5}.*0482

*I pointed out before that the second term has a coefficient that is equal to n.*0490

*Something I also pointed out in the last slide is that there is symmetry around the middle term or terms,*0498

*which tells me that, if I figure out these middle coefficients, well, I already know that the first and last coefficients are 1.*0504

*And I know that the second (and, by symmetry, second-to-last) coefficients are equal to n.*0511

*So then, all I would have to figure out is one of these; and I know that the other one is the same.*0519

*Something else to be aware of is that the sum of the exponents in each term is n.*0523

*Let's look right here: for this term, 10a ^{3}b^{2}, if I add up 3 and 2, that is going to give me 5.*0528

*So, the sum of the exponents is equal to 5.*0537

*Or looking at this, 10a ^{2}b^{3}, I get the same thing.*0540

*The sum of the exponents of each term is equal to n.*0544

*The next concept that we will learn requires that you remember how to work with factorials.*0552

*So, I am going to go ahead and review factorials: factorials are products, and the symbol that looks like an exclamation point is actually read as "factorial."*0557

*This is a product: n! is equal to n times n - 1 times n - 2 times n -3...and you continue on that way until you get all the way to 2, and then finally 1.*0567

*For example, 4 factorial: here, n = 4; this is going to be equal to 4, so here n = 4; n - 1...4 - 1 is 3, so that is 4 times 3.*0586

*4 - 2 gives me 2; 4 - 3 gives me 1; and this equals...4 times 3 is 12, times 2, times 1, is just 24.*0599

*So, 4! is just the product of 4 times 3 times 2 times 1.*0613

*If you are working with fractions that involve factorials in the numerator and denominator, you can often do a lot of canceling out to make things easier.*0618

*If you were working with 8!/6!,that gives you 8 times 7 times 6 times 5 times 4, and then on down to 1.*0625

*In the denominator, you have 6 times 5 times 4 times 3 times 2 times 1.*0637

*So, instead of multiplying all of this out and dividing, it makes a lot more sense to start canceling out all of these common terms.*0644

*1 through 6; 1 through 6; this just leaves me with 8 times 7, or 56.*0657

*Now, the reason we reviewed it is to work with this formula.*0665

*First, we are talking about the binomial theorem, and then the binomial formula.*0668

*The binomial theorem is really what we just discussed; and it is the idea that, when you expand (a + b) ^{n},*0672

*you are going to get an equation with the properties we just discussed.*0682

*And those properties are that the first term is going to be a ^{n};*0688

*the second term is going to have some coefficient; and it is going to be raised to the power a ^{n - 1}.*0692

*b is going to be to the first power; then you are going to get another term; and you are going to have a total of n terms.*0699

*You are going to have another term; it is going to have a coefficient that is going to be a ^{n - 2}.*0707

*b is going b ^{2}; it is going to increase by 1.*0711

*Then, you are going to get another coefficient, a ^{n - 3}; b is going to become b^{3}.*0714

*And that is going to go on and continue on until you get ab ^{n - 1}, b^{n}, and so on.*0721

*Excuse me, plus...that is a ^{n - 1}, times b, plus b^{n}; the last term will be b^{n}; the first term is a^{n}.*0731

*Each of these terms can be given in the general form a ^{n - k}b^{k}.*0751

*And we will look at some examples in a few minutes, and talk about what this means, as we relate it to the binomial formula.*0760

*This is the binomial formula; this is pronounced "n choose k."*0769

*Now, why are we even looking at this--what is this for?*0781

*Think about how we can find these coefficients.*0785

*We can use Pascal's triangle, but that could be really impractical and quite a bit of work,*0788

*if you have to go through that whole large array, if you have a really big expansion.*0793

*And if you just find the coefficient for a particular term, you are going to have to go through a lot of work to get there.*0799

*So, as usual, we have a formula that gets us directly there, without having to go through all that work.*0804

*This formula will give me the coefficients for a term that has the exponent a ^{n - k}b^{k}.*0812

*So, the binomial formula allows me to find this coefficient.*0823

*Let's use an example of (a + b) ^{7}: I can find the coefficient, using this, of a particular term;*0829

*or I can expand the whole thing and find all of the terms.*0839

*Let's look at what this expansion would look like: it would look like this.*0841

*I know that the first term is going to be a ^{7}; recall that the second coefficient is going to be equal to n.*0845

*So this one I also know; it is 7; and then, I know that a...the exponent is going to decrease by 1; and b is going to appear, raised to the first power.*0851

*I also know that I am going to have 8 terms, because I am going to have n + 1 terms, or 7 + 1.*0866

*So, I have another term here; I don't know the coefficient, but I do know that it is going to be a ^{5}b^{2}.*0872

*I have another term with an unknown coefficient: a ^{4}b^{3}.*0878

*I have another term, again with an unknown coefficient: a ^{3}b^{4}.*0884

*Another term is a ^{2}b^{5}; another term is ab^{6}; plus b^{7}.*0890

*I have 8 terms, so I am going to have these two middle terms the same; these two terms will be the same;*0900

*the second-to-last terms will be the same; so I know this is 7; and these two outer terms will be the same, 1 and 1.*0907

*I may be asked to find this expansion; I found this much; now, in order to finish it out, I need to find this coefficient and this coefficient.*0917

*If I have those, I have the rest, because I would have the first half, and then I just reflect for the second half.*0927

*Let's use this formula to find the coefficient for this third term.*0935

*In order to use this formula, I need to know two things: I need to know n, and I need to know k.*0950

*n is easy--we know n equals 7; k...if you look up here, k is the exponent of b for that term.*0956

*So, I am looking for the coefficient that goes right here, where n is 7 and k is 2.*0969

*Once I have that, it is a matter of using the formula: n!...this is going to be 7!/k!, which is 2!...times (7 - 2)!;*0978

*this equals 7! divided by 2!, times 5!.*0995

*I am going down here to complete this: 7 times 6 times 5 times 4 times 3, and so on.*1002

*In the denominator, I have 2!, 2 times 1, times 5 times 4 times 3 times 2 times 1 (factorials).*1010

*Again, I am going to cancel out common factors to make this much easier to work with.*1023

*This leaves me with 7 times 6, divided by 2; that is simply 42/2, or 21.*1030

*Therefore, the coefficient for the third term is actually 21.*1043

*Since I know that, and I know that we are going 1, 7, 21, starting from this end*1052

*(we are also going 1, 7, 21), I could do the same thing to find the fourth term coefficient.*1057

*The fourth term coefficient would have, again, an n equal to 7; but this time, k would be equal to 3.*1067

*Note one thing: instead of having to write all this out, to figure out what k is and everything,*1077

*you can just be aware that the term number equals k + 1.*1082

*Let's look at the fourth term: the term number is 4--that equals k + 1; k = 3.*1092

*For the fourth term, k = 3; so that is 1, 2, 3, 4; I know n = 7; and since this is the fourth term, I quickly know that k = 3.*1101

*So, you can use the binomial theorem, and specifically the binomial formula, to find a specific term of the expanded form of (a + b) ^{n}.*1117

*And I talked, in the last section, about how you can use that formula to find the whole expansion.*1126

*Or you may just need to find a certain coefficient within the expansion.*1133

*Recall that the binomial formula is equal to n!, divided by k!, times (n - k)!, where n is this n,*1140

*and k is the value of the exponent for the b in the term that you are looking for the coefficient of.*1158

*As an example, we are going to do something slightly more complex.*1167

*Instead of just a + b, we are actually going to say a - 2b; and we are going to expand that--it is going to be to the fifth.*1169

*So, we have to account for this -2--that the second term here is not just b; it is -2b.*1178

*I know that the first term is going to be a ^{5}.*1188

*I know that the second term is going to have a coefficient of n, so it is a coefficient of 5, and that a is going to decrease by 1.*1193

*I have to be very careful and not just write b; I have to write -2b.*1200

*Then, I have an unknown coefficient; but I do know that I am going to get a ^{3}, because the exponent that a is raised to will decrease by 1.*1206

*And here, I just have to say that b is really -2b; that is going to be squared; it is going to increase by 1.*1216

*I have another unknown coefficient, and it becomes a ^{3} and goes down to a^{2}; -2b squared goes up to -2b cubed.*1226

*Then, there is another coefficient here, and I get a ^{3}, -2b to the fourth, and then finally...*1237

*actually, this is just a; it goes from a ^{2} to a; and then finally, a is gone, and I have -2b^{5}.*1247

*By symmetry, I know the coefficient of these outer terms, right here, is actually 1; you can think of that as 1.*1256

*The two outer terms, then, have coefficients of 5.*1266

*All right, so back to finding a specific term: let's say that I want to find the coefficient for this third term.*1270

*n is going to equal 5; so in order to find the coefficient for this third term, I need to use the binomial formula; I need to have n, and I need to have k.*1298

*k...I could just say it is one less than the term number, so k is 2.*1307

*Or I could look up here and say it is whatever power this is raised to, which is 2.*1311

*Next, I insert these values into the formula; that is going to give me 5!, divided by 2!, times (n - 2)!.*1318

*This is equal to 5! divided by 2!, times 3!, equals 5 times 4 times 3 times 2 times 1, divided by 2! (2 times 1), times 3 times 2 times 1.*1334

*Cancel out common factors; that leaves me with 5 times 4, divided by 2 times 1, or 20 divided by 2, is 10.*1354

*Therefore, coefficient for the third term is equal to 10.*1368

*I can go back up and put this in here, but there is some simplifying that needs to be done.*1374

*So, the third term equals 10a ^{3}, (-2b)^{2}; this equals 10a^{3}...-2 times -2 is 4b^{2}.*1379

*This equals 10 times 4, a ^{3}, b^{2}, which equals 40a^{3}b^{2}.*1398

*It is a little bit more complicated than just putting the 10 in there.*1408

*And then, something to be careful of, as well: then I can say that by symmetry, this is also 10.*1411

*But I need to work with what is in here, which is actually a ^{2}(-2b)^{3}, so it is somewhat different.*1416

*So, let's look at what this fourth term is: 1, 2, 3,...the fourth term.*1425

*Again, I am going to use that coefficient, 10, but here I have a ^{2}(-2b)^{3}.*1431

*So, it is 10a ^{2}...-2 times -2 is 4, times -2 is -8; so it is times -8b^{3}.*1439

*I can rewrite this as 10 times -8a ^{2}b^{3} = -80a^{2}b^{3}.*1454

*When I simplify that fourth term, this is what it is going to look like.*1466

*That was using the binomial formula to find a specific term of the expanded form.*1471

*In the first example, we need to expand (x + 2) ^{4}.*1481

*The general form is (a + b) ^{n}; here, I do the same thing, just realizing that a is equal to x, and b is equal to 2.*1486

*I am going to get something that looks like this: (a + 2) ^{4}; I am going to have five terms.*1497

*I am going to have n equal 4; so, there are going to be 5 terms.*1502

*The first term is going to be a ^{2}; in this case, it is x, so it is x^{2}.*1508

*Excuse me, not squared; to the n--it is going to be a to the n, which is 4, so it is actually going to be x ^{4}.*1517

*In the next term, I have the coefficient; and the coefficient of the second term is equal to n.*1525

*So, that is going to be 4, times x...this exponent is going to decrease by 1...times 2.*1531

*The next term: I am going to have some unknown coefficient; x is going to go from cubed to squared;*1543

*2 is going to go from being the first power to the second power.*1555

*I am going to have another term, and then x ^{2} is just going to become x; 2 is going to go from being squared to being cubed;*1560

*and then, finally, x drops out, and I get, for my last term, 2 ^{n}, or 2^{4}.*1572

*By symmetry, I know that I am going to have a middle term (I have an odd number of terms--I have one middle term);*1581

*the two next to it are going to have the same coefficient--it is going to be 4.*1587

*I can simplify this later on, because I am going to have to deal with this 2 ^{3} and this 2, and multiply that times the coefficient.*1591

*But right now, let's just leave it like this.*1600

*I am almost done with my expansion; for this middle term, there are two ways I could go, because this is not that large of an expansion.*1603

*I could use the binomial formula, and we are going to get more practice with that in a minute.*1612

*But this time, I am actually going to use Pascal's triangle.*1615

*Remember that it starts out with (a + b) ^{0}, and you end up with 1 as the only coefficient, because this equals 1.*1617

*Then, I get (a + b) ^{1} = (a + b); that gives me 1 and 1 as coefficients.*1632

*Once I have those two, I can find the rest, because I know I have 1's out here.*1639

*And to find the other terms, I add the two above: 1 + 1 gives me 2.*1645

*So, for (a + b) ^{2}, these are my coefficients; then I want (a + b)^{3}.*1651

*And what I am finally looking for, to find this coefficient, is when (a + b) is expanded to the fourth.*1659

*So, I have a 1 out here and a 1 out here; 1 + 2 gives me 3; 2 + 1 gives me 3;*1667

*so the coefficients for this expansion (that is going to have 4 terms) are going to be 1, 3, 3, 1.*1674

*Finally, the one I am looking for: I have a 1 and a 1, which I have up here;*1680

*1 + 3 is 4; 1 + 3 is 4, which I already knew; this gives me my middle coefficient of 6.*1685

*Now, I could have used the binomial formula to find this; but I wanted*1695

*to just do it a little bit differently this time, since it was a fairly small expansion, and use Pascal's triangle.*1699

*Now, I have the expansion, but it is not really simplified; we need to simplify these various terms.*1706

*So, let's take...the first term is already simplified; the second term equals (here is the second term) 4 times x ^{3} times 2; that equals 8x^{3}.*1712

*The third term equals (that is 1, 2...the third term) 6x ^{2}, 2^{2}, equals 6x^{2} times 4, equals 24x^{2}.*1733

*That is the third term; the fourth term equals 4x times 2 cubed; that is 4x times 8; 8 times 4 is 32, so that is 32x.*1750

*The fifth term equals 2 ^{4}; 2^{3} is 8, times 2 again gives me 16.*1764

*Rewriting it all here, the expansion of (x + 2) ^{4} is*1774

*x ^{4} + (second term) 8x^{3} + (third term) 24x^{2} + (fourth term) 42x + (the final term) 16.*1779

*Again, I was able to expand this using my properties that I am aware of for the expansion of a binomial*1794

*to find what these exponents are, how many terms there are going to be, and what this coefficient and this coefficient are.*1802

*I was left with just finding one coefficient that I could have used Pascal's triangle for.*1810

*That is what I did; I found the 6 (or I could have used the binomial formula).*1814

*Once I had these coefficients, all I did was to simplify; and this is the expansion, right down here.*1818

*Here, I am asked to find the fourth term of the expansion of (a + 2b) ^{8}.*1829

*So, I am asked to find this specific term; and we know that we can use the binomial formula, n!/k! times (n - k)!.*1834

*I know that I have n; n = 4; what is k?*1847

*Well, the term number equals k + 1; the term number is 4, so this means 4 = k + 1.*1853

*This gives me k = 3; so I have n = 4; k = 3; I can find the coefficient.*1866

*I need to find the coefficient of this fourth term; it is going to have a coefficient.*1875

*For the first term, it is going to be a raised to some power; for the second component of it, it is going to be 2b (not just b), raised to some power.*1884

*How do I know what these powers are?*1902

*Well, recall that it is going to be equal to (let's give ourselves some more room)...*1904

*when you find a term used in a binomial formula, you also know that it is in the form n - k,*1910

*and then the second part is raised to the k power, because that is what k is--*1919

*it is whatever you are raising b to; and in this case, b = 2b.*1926

*I also know that the sum of the exponents has to equal n; the sum of the exponents of a term equals n.*1934

*This makes sense, because n - k + k needs to equal n; the sum of the exponents must equal n, and that works out; so this makes sense.*1945

*So, the fourth term: I need to find the coefficient using the binomial formula.*1954

*I know what the power a is going to be raised to is, n - k; and I know that the power that 2b is going to be raised to is k.*1960

*Let's go ahead and find the coefficient: the coefficient of the fourth term is going to be equal to n!, which is 4!...*1967

*excuse me, n is actually 8--a correction on that: we are finding the fourth term, but n is right here, so n is actually 8.*1981

*So, it is n! (that is 8!), divided by k! (that is 3!); n (which is 8), minus 3, factorial, equals 8!/3! times 5!.*1993

*That gives me 8! divided by 3 times 2 times 1, times 5!.*2013

*I do my canceling to make things easier to work with.*2026

*This leaves me with 8 times 7 times 6, divided by 3 times 2 times 1.*2032

*6 is equal to 3 times 2, so I have more common factors and more canceling.*2040

*That just leaves me with 8 times 7, divided by 1, which is 56; that was the hardest part here--finding the coefficient.*2045

*The coefficient of this fourth term is equal to 56.*2053

*To write that out completely, I am going to write it out as 56a ^{n - k}; that is 8 - 3;*2058

*times (2b) raised to the k power (k is 3); this equals 56a ^{5}(2b)^{3}.*2071

*Simplifying further: 56a ^{5}...2 cubed is 8, times b^{3}.*2087

*If you multiply out 56 times 8, you will get 448; a ^{5}, b^{3}.*2095

*This is the fourth term of this expanded form of (a + 2b) ^{8}.*2103

*It is a little bit complicated: the first thing I had to do is find the coefficient.*2111

*And I did that by using my binomial formula and knowing that n = 8 and k = the term number, plus 1.*2115

*The term number plus 1 is k, so k is one less than the term number: k is 3.*2128

*I was able to find, then, using this formula, that the coefficient is 56.*2133

*The second thing I had to do is figure out what power I should raise a to, and what power I should raise 2b to.*2139

*Well, when I am using the binomial formula, I am finding the coefficient for a term in this form, which is a ^{n - k}2b^{k}.*2145

*So, I knew my n and my k; so I put my coefficient here and said, "OK, I have a ^{8 - 3}(2b)^{3}," using this.*2153

*Simplify, and then multiply to get 448a ^{5}b^{3}.*2163

*We are asked to expand (2x + 3y) ^{5}...this binomial to the fifth.*2172

*And the general form is (a + b) ^{n}.*2178

*So, here we actually have that a is equal to 2x, and b is equal to 3y; so that makes things a little bit more complicated at the end.*2181

*We have to take these into account when we simplify our final answer.*2190

*Since this is the fifth power, I have an n equal to 5.*2197

*I am going to have a total of 6 terms.*2202

*It is going to look something like this at the end: (2x + 3y) ^{5} =...my first term is 2x raised to the n power,*2205

*which is the fifth; plus the coefficient, and then 2x is going to decrease to the fourth power, and I am going to have my 3y here.*2214

*Plus...some other coefficient; 2x becomes cubed, and 3y increases from an exponent of 1 to an exponent of 2.*2225

*Another coefficient is going to bring me to 2x ^{2}3y^{3};*2241

*another coefficient...now I just have 2x, and 3y ^{4}; and then, finally, 3y^{5}.*2250

*And I expect to have 6 terms--let's verify that I am not missing anything: 1, 2, 3, 4, 5, 6.*2261

*Down here at the bottom, I am going to start building up what our final answer is going to look like, because I have some simplifying that I need to do.*2270

*The first term, I can figure out: I already know that it is 2x ^{5}.*2281

*2 ^{3} is 8, times 2 is 16, times 2 is 32; so that is 32x^{5}.*2286

*The second term: recall that the coefficient for the second term is equal to n--that is 5.*2293

*Also, remember that symmetry is involved: since there are 6 terms (that is an even number of terms), I am going to have two middle terms.*2301

*These two are going to have the same coefficient; the two next to those will have the same coefficients; and the two outer have the same coefficients.*2309

*It is not the same value, once I simplify them; but the same coefficient.*2319

*So, since these two are the same, and these two are the same, this term also has a coefficient of 5.*2323

*Let's work this one out up here; this is the second term, and I know that it is equal to 5 times (2x) ^{4} times 3y.*2332

*That is...2 times 2 is 4; 2 cubed is 8, times 2 is 16; so that is times 16x ^{4}, times 3y.*2346

*Multiplying 5 times 16 times 3 is going to give you 240x ^{4}y; I am going to put that in here; that is 240x^{4}y.*2358

*I don't know what I have for these two middle ones, but it is going to be something here, and it is going to be x ^{3}y^{2}.*2373

*Something is here, then x ^{2}y^{3}; I can figure this one out:*2384

*right here, this is the third term; it equals 5 times 2x, times (3y) ^{4}.*2394

*5 times 2x is just 10x; 3 times 3 is 9, times 3 is 27, times 3 is going to give me 81.*2406

*So, that is times 81y ^{4}; so this one is easy, because it is just 81 times 10, is 810xy^{4}.*2419

*So, that was...actually not the third term: correction--we don't know the third term; the first, second, third, and fourth...*2436

*the two middle terms we don't know; we do know the fifth term--this is the fifth term.*2448

*All right, so the fifth term I can fill in: 810xy ^{4}.*2456

*The sixth term, right here, I can also fill in: the sixth term equals (3y) ^{5}.*2465

*3 ^{3} is 27; 3^{4} is 81; 3^{5}, if you figure it out, is 243; 243y^{5}.*2475

*Just using the properties of binomial expansions, I got most of the terms.*2489

*And since there is symmetry, the two middle terms will have the same coefficient inserted up here.*2496

*And then, I can work out the rest and finalize it.*2505

*So, I only need to find one coefficient, using the binomial formula.*2507

*The binomial formula, recall, is n!/k!, times (n - k)!.*2511

*I know that n = 5; for the first, second, third term, k is equal to 2; you could just say "Third term: 3 - 1 is 2";*2521

*or you could look up here and say, "OK, 3y is raised to the second power, so k = 2."*2532

*Therefore, I am going to have 5!/2!, times (5 - 2)!; this equals 5!/2! times 3!,*2537

*which equals 5 times 4 times 3 times 2 times 1; 2 times 1; 3 times 2 times 1.*2559

*1, 2, and 3 all cancel; this gives me 5 times 4, divided by 2; that is 20/2, or 10.*2568

*So, the missing coefficient here is 10; by symmetry, the missing coefficient right here is also 10; I am almost done.*2578

*So, for the first, second, third term, I have my coefficient; but I need to simplify that a bit.*2588

*My third term equals 10 times (2x) ^{3} times (3y)^{2}.*2595

*This is going to be equal to 10 times...2 cubed is 8x ^{3}; 3 squared is 9y^{2}.*2605

*This is going to give me 80 times 9x ^{3}y^{2}, which is just 720x^{3}y^{2}.*2617

*That is my third term, so I am going to put a 720 right here.*2625

*The fourth term: I also have a 10 right here.*2630

*What is slightly different, though: it is 10 times (2x) ^{2}(3y)^{3}.*2635

*This is going to give me 10 times 4x ^{2}; 3 cubed is 27, so it is y^{3}.*2642

*This is 40 times 27...let's put all of the constants together...x ^{2}y^{3}.*2651

*40 times 27 actually comes out to 1080; you can multiply that out for yourself; x ^{2}y^{3}.*2663

*The fourth term, therefore: I am going to put a 1080 right here.*2673

*So, this is quite a bit of work; but we actually only needed to find one coefficient using the binomial formula,*2676

*because by symmetry, these two middle ones were the same (the two middle coefficients that we will put in the blank);*2682

*and I knew that the second and the second-to-the-last had coefficients equal to n.*2690

*Once I found that a 10 went into these blanks, all I needed to do was work out what each of these terms would simplify to,*2695

*to end up with the expansion that I found down here.*2703

*We are asked to find a specific term this time, not the entire expansion.*2710

*I just want to find the fourth term of (x - 3y) ^{7}, using the binomial formula, n!/k!, times (n - k)!.*2713

*All right, n = 7; what does k equal? Well, it is the fourth term; k is going to be one less than that; 4 - 1...I am going to have k = 3.*2725

*This is going to give me the coefficient for a term in this form.*2737

*It is going to have a coefficient, and then it is going to have x raised to the n - k power, times -3y, raised to the k power.*2741

*This is going to become x ^{7 - 3}, (-3y)^{3}.*2757

*Therefore, I am going to get some coefficient (that I am going to find in a moment), times x ^{4}, times (-3y)^{3}.*2768

*Let's go ahead and find the coefficient: this missing coefficient equals n!, 7!, divided by k!, times (7 - 3)!;*2775

*that equals 7!/3!, times 4!, equals (let's write this out)...divided by 3 times 2 times 1, times 4 times 3 times 2 times 1.*2793

*Do some canceling out: 1's, 2's, 3's, and 4's are the same.*2812

*This leaves me with 7 times 6 times 5, divided by 3 times 2 times 1.*2819

*6 has common factors here of 3 and 2; I can just cancel those out.*2825

*So, this is 7 times 5, divided by 1; so just 7 times 5 equals 35.*2832

*I know that my coefficient is, then, 21 that is what is going to go in this blank.*2838

*So, I am going to end up with...the fourth term is going to be 35 (I am going to put 35 right here in this blank)*2842

*x ^{4}(-3y)^{3}, which equals 35x^{4}...-3 squared is 9, times -3 again is -27.*2849

*So, it is -27y ^{3}; so it is time to get out your calculators and figure out 35 times -27, or multiply it out, if necessary.*2864

*And you will find that that equals -945x ^{4}y^{3}.*2875

*OK, we found the fourth term of this expansion by using the binomial formula to find the coefficient,*2881

*and also, an awareness that the exponent of this x is going to be n - k; the exponent for the -3y is going to be k.*2887

*We figured out the coefficient (it is 35), and then simplified, using our rules for working with exponents.*2896

*That concludes this session on the binomial theorem here at Educator.com; thanks for visiting!*2905

0 answers

Post by Suhani Pant on July 27, 2013

Thank you so much for this amazing series on Algebra II. I greatly appreciate your hard-work and effort.

1 answer

Last reply by: Dr Carleen Eaton

Mon Apr 8, 2013 4:35 PM

Post by Kenneth Montfort on March 7, 2013

You are super great teacher - thank you so much for making algebra 2 super understandable :) it was an awesome course.

1 answer

Last reply by: Dr Carleen Eaton

Sat Jan 5, 2013 9:55 AM

Post by Angel La Fayette on January 2, 2013

The "2" was left out @ 38:13.

0 answers

Post by Timothy miranda on July 3, 2010

Thanks for all the help! This was a great course!