For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Sequences

**Main definitions and theorem:**

Let {*a _{n}*} be a sequence.

**Definitions:**

{

*a*} is_{n}__monotonically increasing__means*a*_{n}_{+1}≥*a*for all_{n}*n*.{

*a*} is_{n}__monotonically decreasing__means*a*_{n}_{+1}≤*a*for all_{n}*n*.{

*a*} is_{n}__monotonic__means {*a*} is one or the other._{n}{

*a*} is_{n}__bounded__means there exists*M*such that |*a*| ≤_{n}*M*for all*n*.

**Theorem:** If {*a _{n}*} is bounded and monotonic, then it converges.

**Hints and tips:**

In determining whether a sequence converges or diverges and what it converges or diverges to, you can ignore the first few terms. What is important is what the later terms do.

Don’t try to analyze sequences by plugging numbers into a calculator. This is extremely unreliable.

You often use the “bounded monotonic” theorem to analyze sequences that are defined

__recursively__, that is,*a*_{n}_{+1}is defined in terms of*a*._{n}When analyzing fractions, focus on the biggest term in the top and the biggest term in the bottom. Remember the following ranking of functions:

You can use L’Hôpital’s Rule for situations of the form or . If you have 0 · ∞, you must put it into fraction form before you can use L’Hôpital. The same goes for 1

^{∞}: Write . Then you can often sort out the exponent using L’Hôpital.When you have an expression of the form

*a − b*, especially when square roots are involved, it is often useful to multiply by to take advantage of the identity (*a − b*)(*a + b*) =*a*² −*b*² .

### Sequences

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
- Definition and Theorem 0:05
- Monotonically Increasing
- Monotonically Decreasing
- Monotonic
- Bounded
- Theorem
- Lecture Example 1 1:31
- Lecture Example 2 11:06
- Lecture Example 3 14:03
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Sequences

*We are here to try some more examples of finding limits of sequences.*0000

*The first one is the sqrt(n ^{2} + 6n) - n.*0006

*There is a trick here when you see square roots and you see them being added and subtracted with things.*0014

*The trick is to multiply the top and bottom by the conjugate.*0023

*What I mean by the conjugate is if you have an expression like a + b,*0040

*You multiply it by a - b/a - b.*0046

*So, if you have a + b, you multiply it by a - b,*0052

*If you have a - b, you multiply it by a + b.*0058

*The point of doing that is to take advantage of this difference of squares formula.*0062

*Remember a + b × a - b = a ^{2} - b^{2}.*0067

*What that does is it squares out the expressions for you and that can often simply a square root formula.*0075

*Let us try that out with this example.*0083

*We have sqrt(n ^{2} + 6n) - n.*0088

*I want to multiply that by its conjugate, the sqrt(n ^{2}+6n)+n.*0094

*To pay for that multiplication, I have to do some division by the same expression,*0104

*sqrt(n ^{2}+6n)+n.*0110

*The point of that is that in the numerator we get this a ^{2}-b^{2} expression.*0115

*We get the sqrt(n ^{2}+6n)^{2}, so that is just n^{2}+6n.*0124

*Minus n ^{2}.*0134

*Then that same denominator.*0140

*sqrt(n ^{2}+6n) + n.*0142

*This simplifies nicely down to 6n/sqrt(n ^{2}+6n) + n.*0150

*We are not done yet with this because we still have something that is not so obvious.*0160

*We have 6n in the numerator, some square root stuff in the denominator.*0167

*The key thing to remember is to go through and look at the top and bottom and see what the biggest thing is in both the numerator and the denominator.*0171

*In the top I see the biggest thing here is n.*0180

*In the bottom, well I see sqrt(n ^{2}), that is approximately n.*0186

*You know there is a 6n next to it.*0190

*The n ^{2} is bigger than the 6n, so the sqrt(n^{2}) is appromixately n.*0194

*Then of course we have an n in the denominator.*0200

*The biggest thing top and bottom would be n.*0204

*What I want to do is divide top and bottom by n.*0206

*That will give me just 6 in the numerator and then in the denominator,*0213

*sqrt(n ^{2}+6n)/n + 1.*0220

*Sqrt(n ^{2}+6n), I can write that as,*0230

*Well I can make that a big square root.*0236

*n ^{2}+6n.*0241

*Now if I want to make this n part of the square root, I have to change it to n ^{2}.*0244

*That in turn becomes 1+6/n.*0248

*This whole thing becomes 6/sqrt(1+6/n)+1.*0258

*Now it is in a format that is very conducive to letting n go to infinity.*0269

*As n goes to infinity, the 6/n goes to 0.*0275

*We get 6/sqrt(1) which is just 1 + 1.*0278

*So, 6/2 and we get our answer, that the sequence converges to 3.*0283

*There are a couple clever idea in that one.*0291

*The first was that we had a square root of something minus something.*0296

*The trick to handling that is by multiplying that by its conjugate.*0301

*We multiplied it by the same expression with a plus there.*0306

*To pay for that multiplication, we had to do a division.*0311

*The point of multiplying by the conjugate,*0315

*Is we can exploit this difference of squares formula.*0319

*We get the sqrt(n ^{2}+6n)^{2}, that gets rid of the square root there.*0324

*We get n ^{2}, and we get this difference of squares formula,*0333

*Which simplifies down into 6n.*0339

*That was the first trick, multiplying by the conjugate.*0342

*The second trick was looking at the expression we got, and then dividing top and bottom by the biggest term that we saw.*0346

*Then recognizing that when we had the square root of n ^{2}, that was really approximately n,*0350

*The biggest term in top and bottom was n.*0360

*We divide top and bottom by n.*0362

*Then do some algebra to simplify it, and we finally get our answer.*0366

*Our final example here is to calculate the limit of 1 + 1/n ^{2n}.*0000

*This is an example that is quite common in Calculus homework sets.*0007

*It is also one that students often make mistakes on.*0013

*People think that as n goes to infinity, 1+1/n ought to go to 1 + 0.*0016

*So, you are 1+0 ^{infinity}, so that is 1^{infinity},*0026

*And people think that goes to 1.*0035

*That is a very common response that students give to problems of this form.*0038

*That is flat wrong.*0043

*Let me emphatically warn you away from making that mistake.*0046

*This is actually something a lot more subtle.*0053

*If you see an expression that looks like 1 to the infinity,*0056

*There is something deeper going on there.*0060

*We are actually going to need some more sophisticated tools for that.*0063

*The trick here is when you have an expression of the form a ^{b},*0070

*Often, a very useful way to write that is,*0078

*Write it in the form, e ^{ln}(a^{b}).*0087

*The point of that is that the ln(a ^{b}),*0097

*Is the same as b ^{ln(a)}.*0100

*Natural log is a way of separating an exponent out to the side of an expression,*0104

*So that it is no longer an exponent.*0113

*Then we have to put an e in there to cancel off the natural log.*0117

*The e is kind of paying for the natural log,*0120

*While the natural log is doing the actual work of moving the exponent out of the exponent,*0123

*So that we just have a multiplication.*0127

*Then, what you often do is you look at the exponent,*0131

*Work on the exponent,*0138

*Now, we have two things multiplied by each other.*0147

*Often, that turns out to be a 0 × infinity situation.*0152

*Which is something that can be manipulated into a l'Hopital's rule situation.*0158

*It is often not l'Hopital's rule immediately, but you can manipulate it into 0/0 or infinity/infinity.*0165

*Then you can use l'Hopital's rule.*0172

*Let us try that out with this example.*0175

*We are going to write it as e ^{ln}(1+1/n)^{2n}.*0177

*That is e ^{2n}, we can pull the exponent out, × ln(1+1/n).*0186

*Now, I am just going to look at the exponent, and we will come back and deal with the e part later.*0195

*2n ln(1+1/n),*0201

*Notice that 2n, 2n goes to infinity, as n goes to infinity.*0205

*ln(1+1/n), well 1/n goes to 0.*0215

*So we get ln(1), ln(1) = 0.*0221

*So this is an infinity × 0 situation.*0228

*Not quite ready for l'Hopital's rule yet, but we can write it as.*0230

*2 ln(1+1/n)/1/n.*0238

*What I did was I took this and, and I pulled it down into the denominator.*0246

*Now, remember ln(1+1/n) goes to 0, and 1/n itself goes to 0.*0253

*So what we have is a 0 over 0 situation,*0258

*And that is something where it is legitimate to use l'Hopital's rule.*0262

*So, we are going to do l'Hopital's rule,*0270

*Which means we are going to take the derivative of both the top and bottom separately,*0276

*Then we find the limit of what we get.*0278

*Let us take the derivative of 2 ln(1+1/n).*0281

*Derivative of ln(x) = 1/x.*0287

*This is 2/1+1/n × the derivative of 1 + 1/n by the chain rule.*0290

*That is the derivative of 1 + 1/n is -1/n ^{2}.*0300

*Remember derivative of 1 is 0.*0306

*Derivative of 1/n, that is n ^{-n},*0309

*So its derivative is -1n ^{-2}.*0312

*In the denominator, the derivative of 1/n, is just -1/n ^{2} again.*0315

*This now simplifies, the -1/n squares cancel.*0324

*We get 2/1+1/n.*0328

*Now that is something very easy to take the limit of.*0333

*As n goes to infinity, the 1/n goes to 0, so that is 2.*0337

*We are not finished that, remember this was all the exponent on the e.*0344

*The actual answer is, the limit is e ^{2}.*0348

*That sequence converges to e ^{2}.*0357

*Let us go back to the beginning.*0360

*If you had made this mistake of thinking, oh that is just 1 ^{infinity},*0362

*Which is just 1,*0367

*You would have gotten your answer to just be 1 which is very different from e ^{2}.*0370

*That just shows that it really is a bad mistake.*0373

*The problem is a lot more subtle than that.*0378

*The real secret to this one is first to take e ^{ln} of the expression that you are given.*0381

*Then you sort out the natural log by remembering that you can pull exponents outside of natural logs.*0390

*Then, the expression that you get turns into an infinity × 0 form.*0400

*Infinity × 0 is not ready for l'Hopital's yet,*0405

*But you can rewrite it as 0/0, by pulling n down into the denominator.*0409

*Then it is ready for l'Hopital's rule.*0416

*L'Hopital's rule says you take the derivative of the top and bottom separately.*0420

*After you do l'Hopital's rule, it simplifies a bit, it is an easy limit,*0425

*Then you just had to remember that you are actually working out the limit of the exponent.*0431

*So to find the limit of the whole thing, you had to do e to that power.*0436

*That is the end of our lecture on sequences.*0440

*This is educator.com.*0442

*Hi, this is educator.com and this is the chapter on sequences.*0000

*We are going to be exploring some different ways to find limits of sequences.*0005

*There are several definitions that lead us up to a big theorem that sometimes be a very powerful way to show that a sequence converges.*0011

*Also, to show its limit.*0021

*The definitions we have are monotonically increasing means that the terms of the sequence are getting steadily bigger.*0023

*The a _{n} term is getting steadily bigger in other words the a_{n+1} term is bigger than or equal to the a_{n} term for all n.*0030

*Monotonically decreasing, same idea except the terms are getting steadily smaller.*0040

*The word monotonic just means the sequence is either monotonically increasing, or monotonically decreasing.*0046

*If either one of those is true then you say the sequence is monotonic.*0055

*Bounded means that all of the terms of the sequence means that all of the terms in absolute value are less than some constant number m.*0060

*The big theorem that we are going to use here is that if you can show that a sequence is both bounded and monotonic,*0070

*Meaning that if you can show it is monotonically increasing or decreasing, and that is bounded,*0078

*Then the sequence converges.*0085

*Let us see how that works with an example.*0088

*The example here is a sequence that starts at the sqrt(2).*0091

*Then the way you get successive terms is you take a previous term is by adding 2 and taking its square root.*0095

*For example, a _{1} would be the sqrt(2+a_{0}), so 2 + sqrt(2)*0103

*a _{1} would be the sqrt(2+a_{1}), and so on.*0110

*We have to show that that sequence converges and then we are going to find its limit.*0120

*We will use our definitions and theorem.*0125

*Our first claim here is that the sequence is bounded above by 2.*0130

*In other words, I claim that every term in the sequence is < or = 2.*0155

*To prove this, note that the first term of the sequence, the a a _{0} term,*0165

*Is certainly less than 2 because the sqrt(2) is about 1.4.*0172

*That is certainly less than 2.*0177

*If a _{n} is < 2, if one term < 2,*0182

*Then I am going to a do a little arithmetic here.*0195

*2 + a _{n} would then be less than 4.*0196

*So the sqrt(2+a _{n}) would be less than sqrt(4) which is 2.*0202

*That is saying that a _{n} + 1 < 2.*0208

*If one term is less than term, then the next term is less than 2.*0213

*We already showed that the first term was less than 2.*0220

*So, every term is less than 2.*0225

*Each term forces the next term to be less than 2.*0230

*That proves the claim that the sequence is bounded above by 2.*0236

*That really is an argument by mathematical induction.*0240

*The second claim is that the sequence is monotonic.*0245

*I claim that this sequence is monotonically increasing.*0260

*So I claim that each term is bigger or equal to the previous term.*0270

*The check whether this claim is true, this is true if and only if*0280

*Well to be monotonically increasing, is to say that a _{n} + 1 is bigger than or equal to a_{n}.*0290

*Let us plug in what a _{n} + 1 is.*0300

*By definition that is 2 + sqrt(2 + a _{n}), bigger than or equal to a_{n}.*0305

*I am going to work with this a little bit.*0310

*I am going to square both sides, this is saying 2 + a _{n} is bigger than or equal to a_{n}^{2}.*0312

*If I move all of the terms over to the right, that is saying 0 is bigger than or equal to a _{n}^{2} - a_{n} - 2.*0322

*I can factor that into (a _{n} - 2) × (a_{n} + 1).*0330

*Let us remember that since a _{n} < 2.*0345

*Over on the left we showed that every term is < 2.*0355

*Since a _{n} < 2, a_{n} - 2 < 0.*0360

*Since all of these terms are positive, a _{n} + 1 would be bigger than 0.*0370

*a _{n} - 2 × a_{n} + 1 is less than or equal to 0.*0377

*Which means that this equation is true and this equation is true.*0388

*a _{n} + 1 is > or = a_{n}.*0395

*What we showed here is that the sequence is bounded and monotonically increasing.*0403

*Our theorem kicks in here and says a bounded monotonic series converges.*0410

*We can invoke the theorem.*0419

*By the theorem the sequence converges.*0428

*That shows the sequence converges, so we have done half of the problem there.*0438

*The sequence converges.*0444

*The second half is to find its limit.*0446

*We will do that on the next page.*0447

*To find its limit, we know now that it does have a limit because we showed that it converges on the previous page.*0450

*Let us call that limit L.*0460

*That is saying that a _{n} converges to L.*0465

*There is a little trick here which is that if a _{n} converges to L, then the sequence of successive terms must also converge to L.*0467

*But a _{n} + 1 is the sqrt(2 + a_{n})*0482

*So, a _{n} + 1 = sqrt(2 + a_{n}).*0491

*But, a _{n} converges to L, so sqrt(2 + a_{n}) converges to 2 + L.*0503

*Now we have a _{n} converging to L, but it also converges to 2 + L.*0512

*So sqrt(2 + L), L must be equal to sqrt(2 + L).*0518

*Now we want to solve that equation and figure out what L is.*0527

*L ^{2} = 2 + L, L^{2} - L - 2 = 0.*0533

*(L-2)(L+1) = 0.*0537

*L is either 2, or -1.*0545

*The limit is 1 of those two possible numbers.*0550

*Remember that all of these terms are positive*0555

*So, you cannot have a bunch of positive numbers converging to -1.*0561

*We cannot have a _{n} converging to -1, so a_{n} must converge to the other possible limit, 2.*0572

*Let us recap that problem there.*0589

*We were given this fairly tricky sequence and we had to show that it converges first*0592

*The way we showed it converges was we tried to show that it was bounded and we showed that it was monotonic.*0600

*Monotonically increasing.*0612

*Then those two together allowed us to invoke the theorem which says a bounded monotonic sequence converges.*0614

*Those two conditions allowed us to say that the sequence converges.*0624

*Then once we know it converges we can assume that its limit is L and go through this little algebraic trick looking at what a _{n} + 1 converges to.*0632

*Set those two limits equal to each other because the sequence only converges to one limit.*0646

*Do a little algebra to find the possible limits.*0651

*Then see which limit makes sense.*0658

*The limit that makes sense was 2 and so the limit must be 2.*0662

*Let us try something a little more computational.*0668

*Here we are given the sequence 3n + 5n ^{2}/2n^{2} + 6.*0671

*The trick here is to find the largest term in the numerator and the denominator and divide.*0682

*Here we look at the numerator 3n + 5n ^{2}, definitely the largest term there is 5n^{2}.*0709

*In the denominator, 2n ^{2} + 6, even though the 6 is bigger than the 2, the n^{2} is going to be much bigger in the long run.*0720

*So the 2n is going to be much bigger there.*0730

*In fact, what really matters here is not the coefficients, it is the powers on the n's.*0733

*The fact that we have an n ^{2} in the numerator and the denominator is the important thing.*0740

*What we are going to do is divide top and bottom by n ^{2}.*0745

*When you divide top and bottom by the same number, that is really multiplying all of it by 1.*0753

*That is legitimate and it is not going to change the limit.*0760

*We get (3n/n ^{2} + 5n^{2}/n^{2})/(2n^{2}/n^{2} + 6/n^{2}).*0765

*(3n/n ^{2} is 3/n + 5)/(2 + 6/n^{2}).*0783

*In the limit, as n goes to infinity, the 3/n is going to go to 0.*0793

*The 6/n ^{2} is also going to go to 0 so we end up with 5/2 as our limit.*0799

*To recap when you have one of these fractional situations*0810

*Where you have an expression in the numerator and an expression in the denominator.*0814

*You want to look at both numerator and denominator and find the largest term in sight,*0819

*And divide both top and bottom by that largest term.*0822

*Here the largest term was n ^{2}, so we divide top and bottom by n^{2}.*0828

*That really makes all the other terms go to 0, and it leaves you with the terms that you need to determine the limit.*0834

*Let us try another limit example.*0844

*We want to find the limit of the sequence of sqrt(n)/ln(n).*0845

*Here, both of these sqrt(n) and ln(n), when n goes to infinity, both of these go to infinity.*0853

*What we are really looking at is a situation of something going to infinity/something going to infinity.*0865

*The classic rule to use here that you might remember from Calculus 1 lectures is l'Hopital's Rule.*0873

*Remember, you can use l'Hopital's Rule in two situations.*0889

*You can use it when you have a limit going to infinity/infinity or 0/0.*0894

*If you have 0 × infinity, you cannot use l'Hopital's Rule directly,*0903

*But you try to rewrite it as an infinity over infinity or 0 over 0 situation.*0910

*Then you can use l'Hopital's Rule.*0916

*Let us remember what l'Hopital's Rule does.*0921

*It says you can take the derivative of the top and bottom.*0923

*The strange thing here is you do not take the derivative using the quotient rule you learned in Calculus 1.*0925

*Instead you take the derivative of the top and bottom separately.*0934

*The derivative of the sqrt(n), we think of that as being n ^{1/2}.*0947

*Its derivative is 1/2n ^{-1/2}, that is square root of n in the denominator.*0953

*The derivative of ln(n) is 1/n.*0962

*If we take the denominator and flip it up to the numerator, we get n/2sqrt(n).*0968

*n/sqrt(n) is the sqrt(n).*0977

*Now we want to take the limit as n goes to infinity of sqrt(n/2).*0981

*That clearly goes to infinity.*0997

*We say that this sequence diverges to positive infinity.*1001

*To recap what we did with that example, we were given sqrt(n)/ln(n).*1014

*We noticed both of those go to infinity, so we have a limit that goes to infinity over a limit that goes to infinity.*1021

*That is l'Hopital's Rule.*1026

*l'Hopital's Rule says you take the derivative of the top and bottom and you take those two derivatives separately.*1030

*You do not use the regular quotient rule that you use in Calculus 1.*1036

*We simplify it using a little bit of algebra, and then we talk the limit of the new expression that we get.*1044

*Then that tells us what happens to the original sequence, it diverges to infinity.*1052

*Let us try some more examples later.*1053

*This is educator.com*1054

1 answer

Last reply by: Dr. William Murray

Wed Dec 28, 2016 11:13 AM

Post by Acme Wang on December 27, 2016

Hi Professor,

Just a bit confused about Example II, the question does not say take the limit as n goes to infinity, how could you just assume that? :):) Really nice video! Thank you very much!

1 answer

Last reply by: Dr. William Murray

Wed Aug 19, 2015 10:06 AM

Post by Mitchell Mayberry on August 17, 2015

Are most of the problems in the sequences section for an average Calc 2 class going to be taking the limit?

1 answer

Last reply by: Dr. William Murray

Wed Jul 1, 2015 8:50 AM

Post by Rafael Mojica on June 26, 2015

I dont understand why we went from sequences to taking the limit.

1 answer

Last reply by: Dr. William Murray

Fri Jun 27, 2014 5:02 PM

Post by robert moreno on June 27, 2014

in ex.4 can you use l'hospital's rule after you use the conjugate?

1 answer

Last reply by: Dr. William Murray

Sat Apr 19, 2014 5:37 PM

Post by Taylor Wright on April 16, 2014

In Ex1, how could (an-2)(an+1) be less than or equal to zero if an-2<0 and an+1>0 ? When multiplied together couldn't they only produce a negative value, therefore (an-2)(an+1) should only be < 0 ? I don't see how their product could ever equal zero.

Thank you.

1 answer

Last reply by: Dr. William Murray

Mon Oct 21, 2013 7:45 PM

Post by Nommel Meless Ghislain Djedjero on October 17, 2013

I meant not by 3?

1 answer

Last reply by: Dr. William Murray

Mon Oct 21, 2013 7:45 PM

Post by Nommel Meless Ghislain Djedjero on October 17, 2013

How do you know that the sequence is bounded above by 2 and not by 1 ?

1 answer

Last reply by: Dr. William Murray

Fri Aug 31, 2012 5:39 PM

Post by Brian Raaflaub on July 27, 2012

How do you determine the divergence or convergence of the sequence sin(2n)/(1+(n^1/2)) ??

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:58 AM

Post by Real Schiran on November 4, 2011

Cool !!

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:57 AM

Post by khadar mire on October 25, 2011

very helpful.thanks eductors team

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:54 AM

Post by nizar ayadi on May 1, 2011

How can We determine if

the sequense n^2 e ^ (-n) is convergent or divergent ?

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:50 AM

Post by mary setlock on July 5, 2010

:) i just love BC calculus. what a great way to spend a monday night!

happy integrating!!!!!!

1 answer

Last reply by: Dr. William Murray

Mon May 13, 2013 10:48 AM

Post by Collin Wilson on March 21, 2010

Hello,

in example 2, the notation should (for completeness) should be

lim (n-> inf) 3n+5n^2 / 2n^2 +6

:)