# Math Help - (FP1) summation of finite series using standard results

1. ## (FP1) summation of finite series using standard results

Find the sum of the series:
$1^2-2^2+3^2-4^2+...-(2n)^2$
I do the sum and each time I get a different answers, none of them matching with the answer in my book.
My latest working:
(a) $1^2, 3^2, 5^2, ...(2n-1)^2, so the series can be represented by (2r+1)^2$
starting from r=1, there is (n-1) terms and 1^2 have to be added separately.
(b) $2^2,4^2,6^2,...(2n)^2, so the series can be represented by (2r)^2$
starting from r=1, there is n terms.
----
(a) $1+\sum\limits_{r=1}^{n-1} (2n-1)^2$
$=1+4\sum\limits_{r=1}^{n-1} r^2-4\sum\limits_{r=1}^{n-1} r+\sum\limits_{r=1}^{n-1}1$
$=4/3 n^3-4n^2+11/3 n$
(b) $\sum\limits_{r=1}^{n} (2r)^2
=4/3n^3-2n^2+2/3n$

Summation=(a)-(b)=-6n^2+3n
=-3n(2n-1)
Can you please narrow down the field where I should search the error?
I am exasperated looking through my rough handwriting.

2. Hello,

$1^2-2^2+3^2-\dots+(2n-1)^2-(2n)^2$

(a) ${\color{blue}1}+\sum\limits_{r=1}^{n-1} (2{\color{red}r}-1)^2$
Small modification in red =)
Question in blue : why did you add 1 ???
Other question : it doesn't go 'til n-1, but n :
$[2(n-1)-1]^2=[2n-3]^2 \neq (2n-1)^2$

Try to rethink your series with these remarks

And I guess it's better to leave the summations in their factored form, that is $\sum_{k=1}^n ~ k^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^n ~k = \frac{n(n+1)}{2}$

$\sum_{k=1}^n ~ 1=n$

3. Originally Posted by Moo
Hello,

$1^2-2^2+3^2-\dots+(2n-1)^2-(2n)^2$

Small modification in red =)
Question in blue : why did you add 1 ???
Other question : it doesn't go 'til n-1, but n :
$[2(n-1)-1]^2=[2n-3]^2 \neq (2n-1)^2$

Try to rethink your series with these remarks

And I guess it's better to leave the summations in their factored form, that is $\sum_{k=1}^n ~ k^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^n ~k = \frac{n(n+1)}{2}$

$\sum_{k=1}^n ~ 1=n$
I added 1 because when r=1, the series starts from 3^2, and that extra one means one term less in the summation
Am I right?

I added 1 because when r=1, the series starts from 3^2, and that extra one means one term less in the summation
Am I right?
What is 2r-1 if r=1 ?

5. Originally Posted by Moo
What is 2r-1 if r=1 ?
Stop winking and tell me whether I have a mistake in my working.
(I get it, but my choice was also logical, why would it give a wrong result?
Pinpoint the error while I do the sum your way, ok?)

Stop winking and tell me whether I have a mistake in my working.
(I get it, but my choice was also logical, why would it give a wrong result?
Pinpoint the error while I do the sum your way, ok?)
Uuuh ?
Winking was a kind way of asking it ! So that I don't look like a severe person
There's no problem, I'll point out your mistakes !

Although your choice was logical, it was wrong, because you were repeating some terms while you were forgetting some others :s

7. Originally Posted by Moo
Uuuh ?
Winking was a kind way of asking it ! So that I don't look like a severe person
There's no problem, I'll point out your mistakes !

Although your choice was logical, it was wrong, because you were repeating some terms while you were forgetting some others :s
The winking comment wasn't a rebuke
I throw up my hands in the air, what was I repeating?
Though I admit I was ridiculously convoluted.

8. Okay, I'll try to explain !

You first wrote $1+\sum_{r=1}^{n-1} (2r-1)^2$
But this equals :

$1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2$

This is supposed to equal $1^2+3^2+\dots+(2n-1)^2$
So the red part is repeating !

Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

$1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2$

Similarly, $2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2$

____________________________________
The sum S you're looking for is :

\begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

The blue stuff simplifies to 0.

So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\
&=-4 \cdot \frac{n(n+1)}{2}+n \\
&=-2n(n+1)+n \\
&=-n(2n+1) \end{aligned}

Is it clearer ?

9. Originally Posted by Moo
Okay, I'll try to explain !

You first wrote $1+\sum_{r=1}^{n-1} (2r-1)^2$
But this equals :

$1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2$

This is supposed to equal $1^2+3^2+\dots+(2n-1)^2$
So the red part is repeating !

Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

$1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2$

Similarly, $2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2$

____________________________________
The sum S you're looking for is :

\begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

The blue stuff simplifies to 0.

So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\
&=-4 \cdot \frac{n(n+1)}{2}+n \\
&=-2n(n+1)+n \\
&=-n(2n+1) \end{aligned}

Is it clearer ?
I wrote
$

$
$1+\sum_{r=1}^{n-1} (2r-1)^2
" alt="1+\sum_{r=1}^{n-1} (2r-1)^2
" /> here but did the sum first time with this

$

$
$1+\sum_{r=1}^{n-1} (2r+1)^2
" alt="1+\sum_{r=1}^{n-1} (2r+1)^2
" />
So I retained r=1 in 2r+1=3 and the series goes when I switched to (2r-1)^2 second time when I did the sum.

My head is feeling light, I wonder why

10. Originally Posted by Moo
Okay, I'll try to explain !

You first wrote $1+\sum_{r=1}^{n-1} (2r-1)^2$
But this equals :

$1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2$

This is supposed to equal $1^2+3^2+\dots+(2n-1)^2$
So the red part is repeating !

Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

$1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2$

Similarly, $2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2$

____________________________________
The sum S you're looking for is :

\begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

The blue stuff simplifies to 0.

So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\
&=-4 \cdot \frac{n(n+1)}{2}+n \\
&=-2n(n+1)+n \\
&=-n(2n+1) \end{aligned}

Is it clearer ?
I wrote
$

1+\sum_{r=1}^{n-1} (2r-1)^2
$
here but did the sum first time with this

$

1+\sum_{r=1}^{n-1} (2r+1)^2
$

So I retained r=1 in 2r+1=3 and the series goes when I switched to (2r-1)^2 second time when I did the sum.

My head is feeling light, I wonder why