Prove (2n+1)+(2n+3)+ +(4n-1)=3n^2

[Jef123]

1. Prove (2n+1)+(2n+3)+(2n+5)...+(4n-1)=3n²



3. First, (2(1)+1) = 3(1)²

Now for the inductive step, (2(n+1)+1)+(2(n+1)+3)+...+(4n-1)+(4n+1)+(4n+3) = 3n²+4n+1+4n+3

(2n+3)+(2n+5)+...(4n-1)+(4n+1)+(4n+3) = 3n²+8n+4

This is where I am stuck because I know I need to show that the RHS = 3(n+1)². I figured that I if I subtract (2n-1) on the RHS I will get the answer, but I'm not sure if that would be a mere coincidence and if I am suppose to do that, I do not understand why that step should be made.

 

[verty]

I wouldn't use induction for this. Assume n>=1, now count the terms and treat the 2n's and the odd numbers separately.

 

[1MileCrash]

No, subtracting the 2(n+1) from the series for 3n² does not get you the series for 3(n+1)² because these series are not infinite. Think about it, how could I subtract something from 3n² and get a larger number?

Instead, try noting that a number is added to every term for 3(n+1)².

 

[Jef123]

I thought I might have been able to subtract (2n+1) because when adding the n+1 to the LHS. Doesn't it "shift" the lower and upper limits? Because now the series starts with 2n+3 and not 2n+1 and ends with 4n+3 and not 4n-1. Or is it still suppose to start with 2n+1?

 

[George Jones]

Are you allowed to use the expression for the sum of an arithmetic series? If so, just apply it directly.

 

[D H]

I wouldn't use induction for this. Assume n>=1, now count the terms and treat the 2n's and the odd numbers separately.

Are you allowed to use the expression for the sum of an arithmetic series? If so, just apply it directly.

Given that this is a somewhat odd place to use induction, I suspect that the assignment is to nonetheless use induction.

I figured that I if I subtract (2n-1) on the RHS I will get the answer, but I'm not sure if that would be a mere coincidence and if I am suppose to do that, I do not understand why that step should be made.

As noted, this approach will not work.

Denote the left hand side as S_n. Without using the right hand side (yet), what's the relationship between S_n+1 and S_n? You should get S_n+1 = S_n + other stuff. Yes, the first term now starts with 2n+3 rather than 2n+1, so you can look at -(2n+1) as one term in that "other stuff".

Once you get this relationship resolved, you can use induction step and substitute S_n=3n² in this expression.

 

[LCKurtz]

1. Prove (2n+1)+(2n+3)+(2n+5)...+(4n-1)=3n²
3. First, (2(1)+1) = 3(1)²

Now for the inductive step

To keep things straight, you are given
(2n+1)+(2n+3)+(2n+5)...+(4n-1)=3n²

And you want to prove:

(2n+3)+(2n+5)+...(4n-1)+(4n+1)+(4n+3) = 3n²+8n+4

[Edit - added] Don't you want 3(n+1)² on the right. I didn't notice that at first.
[Edit again] I see DH noticed that about the same time. Nevertheless, the below suggestion applies.

Notice on the left sides, the second sum is missing the first term and has two terms added at the end compared to the first sum. So try building the second equation from the given first equation by subtracting that 2n+1 from both sides of the first equation and adding those two terms on the end to both sides. See what comes out on the right side.

 

[D H]

(2n+3)+(2n+5)+...(4n-1)+(4n+1)+(4n+3) = 3n²+8n+4

The left hand side is not 3(n+1)². Try doing that again.

 

[Jef123]

Okay so... (2n+1)+(2n+3)+(2n+5)+...+(4n-1)+(4n+1)+(4n+3)-(2n+1) = 3n²+(4n+1)+(4n+3)-(2n+1)

Then the RHS would = 3n²+6n+3 = 3(n+1)² as desired.

I think I may get it now. So correct me if I am wrong, but because S_n+1 "loses" 2n+1 at the beginning of the series (on the LHS), then I must subtract it from the end of the series on the LHS. But on the RHS 2n+1 is still part of 3n² so, although I added (4n+1) and (4n+3), that formula still has an additional (2n+1) that should not be there.

sorry if that wasn't explained well, I am still new to this analysis textbook and am not very comfortable with it yet. And as D H mentioned that this is an odd place to use induction, just out of curiosity, what method would be more appropriate to prove this question or these types of questions?

 

[LCKurtz]

Please quote the message to which you are responding. I don't know if you are responding to D H or me.

Okay so... (2n+1)+(2n+3)+(2n+5)+...+(4n-1)+(4n+1)+(4n+3)-(2n+1) = 3n²+(4n+1)+(4n+3)-(2n+1)

Then the RHS would = 3n²+6n+3 = 3(n+1)² as desired.

I think I may get it now. So correct me if I am wrong, but because S_n+1 "loses" 2n+1 at the beginning of the series (on the LHS), then I must subtract it from the end of the series on the LHS. But on the RHS 2n+1 is still part of 3n² so, although I added (4n+1) and (4n+3), that formula still has an additional (2n+1) that should not be there. (Nevermind, I see it is subtracted at the end).

I wouldn't state it that way. All you are trying to do is derive the second (to prove) equation from the first (given). You do whatever steps are necessary to do that. And is that red term supposed to be there? (Nevermind, I see it is subtracted at the other end).

 

[D H]

sorry if that wasn't explained well, I am still new to this analysis textbook and am not very comfortable with it yet. And as D H mentioned that this is an odd place to use induction, just out of curiosity, what method would be more appropriate to prove this question or these types of questions?

I'd do what George said. First express your left hand side as a formal series:
(2n+1)+(2n+3)+\cdots+(4n-1) = \sum_{k=1}^n 2n-1+2k
This is a finite series. We can rearrange terms, split the series into a sum of series, factor out common terms, etc. Aside, but note well: You can't necessarily do these things with an infinite series.
\sum_{k=1}^n 2n-1+2k<br /> = \sum_{k=1}^n (2n-1) + \sum_{k=1}^n 2k<br /> = (2n-1)\sum_{k=1}^n 1 + 2\sum_{k=1}^n k
There are some things you should just know off the top of your head, and these two sums ($\sum_{k=1}^n 1$ and $\sum_{k=1}^n k$) are in that camp because they come up all the time. Or you can look them up, or derive them. What you'll remember/find/derive is that $\sum_{k=1}^n 1 = n$ and $\sum_{k=1}^n k = \frac{n^2+n}2$. Plugging those into the above yields
\sum_{k=1}^n 2n-1+2k = (2n-1)n+2\frac{n^2+n} 2 = 3n^2