# Thread: Sum of the series

1. ## Sum of the series

Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-7^2+.....

2. ## Re: Sum of the series

Originally Posted by blackhat123

Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-7^2+.....

Are you missing a 6^2 term?

3. ## Re: Sum of the series

Ya I missed it out.

Thanks for pointing it.
The problem should be like this
Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-6^2+.....

4. ## Re: Sum of the series

Here's one way it could be solved:

$\displaystyle \sum_{x=1}^{n}(2 x-1)^2 = \frac{1}{3} (4 n^3-n)$
$\displaystyle \sum_{x=1}^{n}(2 x)^2= \frac{2}{3} n (n+1) (2 n+1)$
$\displaystyle \sum_{x=1}^{n}(2 x-1)^2-\sum_{x=1}^{n}(2 x)^2 =\frac{1}{3} (4 n^3-n)-\frac{2}{3} n (n+1) (2 n+1)= -2 n^2-n$

That gives you 2 terms for each n, or 2n terms. I'm not certain this is how it's *supposed* to be solved, though

5. ## Re: Sum of the series

Thanks Stro.So this is the sum of the series upto the first n terms.Now should I replace n by 2n to get the sum of the series upto 2n terms?Please guide me.

6. ## Re: Sum of the series

Edit: To answer your question: No, the above is the summation up to 2n terms within the original series, it would not be changed at this point, but you would have to show your work I bet.

ie. $\displaystyle \sum_{x=1}^{n}(2x)^2=4\cdot\sum_{x=1}^{n}x^2= 4\cdot\frac{n(n+1)(2n+1)}{6}=\frac{2}{3}n(n+1)(2n+ 1)$
Spoiler:
$\displaystyle \sum_{x=1}^{n}(2 x-1)^2$
$\displaystyle \sum_{x=1}^{n}4x^2-4x+1$
$\displaystyle 4\cdot\frac{n(n+1)(2n+1)}{6}-4\cdot\frac{n(n+1)}{2}+n$
$\displaystyle \frac{4n^3+6n^2+2n}{3}-\frac{6n^2+6n}{3}+n$
$\displaystyle \frac{1}{3}(4n^3+6n^2+2n-6n^2-6n+3n)$
$\displaystyle \frac{1}{3} (4 n^3-n)$

7. ## Re: Sum of the series

The sum of squares, for 1 to n, is well known to be (1/6)n(n+1)(2n+1) so the sum of squares from 1 to 2n is (1/6)(2n)(2n+1)(4n+1)= (1/3)n(2n+1)(4n+1).

Here, however, we have a negative sign on all even squares so we have to subtract off $\displaystyle 4+ 16+ ...+ (2n)^2$ twice.
But $\displaystyle 4+ 16+ ...+ (2n)^2= 4(1)+ 4(4)+ ... 4(n^2)= 4(1+ 4+ ...+ n^2)$. Using the formula for sum of squares as before, that is 4(1/6)n(n+1)(2n+1)= (2/3)n(n+1)(2n+1). Notice that this is 1 to n, not 2n, now. Since we have to subtract that off twice, the original sum is
(1/3)n(2n+1)(4n+1)- (4/3)n(n+1)(2n+1)= (1/3)n(2n+1)(4n+1- 4n-4)= (1/3)n(2n+1)(-3)= -n(2n+1).

Check: If n= 3, 1- 4+ 9- 16+ 25- 36= -21
-3(2(3)+1)= -3(7)= -21.

8. ## Re: Sum of the series

Thanks all for helping me out.