1. ## Series

Show that, when $2n$ terms of the series

$1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$

are grouped in pairs, an arithmetic series is formed.

2. Originally Posted by SengNee
Show that, when $2n$ terms of the series

$1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$

are grouped in pairs, an arithmetic series is formed.
Grouping the terms gives pairs of the form:

$u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

so:

$\sum_{k=1}^{2n}(-1)^{k+1}k^2 = \sum_{r=1}^n (1-4r)$

RonL

3. Originally Posted by SengNee
Show that, when $2n$ terms of the series

$1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$

are grouped in pairs, an arithmetic series is formed.
Originally Posted by CaptainBlack
Grouping the terms gives pairs of the form:

$u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

so:

$\sum_{k=1}^{2n}(-1)^{k+1}(2k)^2 = \sum_{r=1}^n (1-4r)$

RonL
So to explicitly address the show bit, now that the hard work's been done:

$u_{r+1} - u_r = .....$ therefore ....

4. Originally Posted by CaptainBlack
Grouping the terms gives pairs of the form:

$u_r=(2r-1)^2-(2r)^2=(2r)^2-4r+1-(2r)^2=1-4r$

so:

$\sum_{k=1}^{2n}(-1)^{k+1}(2k)^2= \sum_{r=1}^n (1-4r)$

RonL
$\sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
I don't understand this part...

5. Originally Posted by SengNee
$\sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
I don't understand this part...
It's a shorthand way of writing $1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$.

If you haven't met this notation before, don't worry about it.

If you want to research it a bit, it's called summation or sigma notation.

6. Hello, SengNee!

Here's a rather primitive approach . . .

Show that, when $2n$ terms of the series

$1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+ \cdots +(2n-1)^2-(2n)^2$

are grouped in pairs, an arithmetic series is formed.

Group in pairs: . $(1^2-2^2) + (3^2-4^2) + (5^2-6^2) + (7^2-8^2) + \cdots + \left([2n-1]^2 - [2n]^2\right)$

. . . . . . . . . $= \quad(-3)\quad+\quad(-7) \;\;\;+\;\;\; (-11) \;\;+\;\; (-15)\quad + \cdots +\quad \left(-[4n-1]\right)$

This is an arithmetic series with first term $a = \text{-}3$ and common difference $d = \text{-}4$

7. Originally Posted by mr fantastic
It's a shorthand way of writing $1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$.

If you haven't met this notation before, don't worry about it.

If you want to research it a bit, it's called summation or sigma notation.
$\sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
$\sum_{k=1}^{2n}(-1)^{k+1}(k)^2$

I think the first one is false, the second one is correct.
Because for the first one, when $k=2n$,

$u_{2n}$
$=(-1)^{2n+1}(4n)^2$
$=-(4n)^2$

For the second one, when $k=2n$,

$u_{2n}$
$=(-1)^{2n+1}(2n)^2$
$=-(2n)^2$

Isn't it?

8. Originally Posted by Soroban
Hello, SengNee!

Here's a rather primitive approach . . .

Group in pairs: . $(1^2-2^2) + (3^2-4^2) + (5^2-6^2) + (7^2-8^2) + \cdots + \left([2n-1]^2 - [2n]^2\right)$

. . . . . . . . . $= \quad(-3)\quad+\quad(-7) \;\;\;+\;\;\; (-11) \;\;+\;\; (-15)\quad + \cdots +\quad \left(-[4n-1]\right)$

This is an arithmetic series with first term $a = \text{-}3$ and common difference $d = \text{-}4$

$S_{2n}=1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$
$S_{2n}=-(2n)^2+(2n-1)^2-(2n-2)^2+(2n-3)^2+...-2^2+1^2$
$2S_{2n}=(1-4n^2)+(4n^2-4n-3)+(5+8n-4n^2)+$ $(4n^2-12n-7)+...+(4n^2-4n-3)+(1-4n^2)$
$2S_{2n}=(-4n-2)+(-4n-2)+...+(-4n-2)$
$2S_{2n}=2n(-4n-2)$
$S_{2n}=n(-4n-2)$
$S_{2n}=-2n(2n+1)$

I do like this. Is it correct also?

9. $Can \quad anyone \quad help \quad me({\color{blue}\text{8th}} \quad post)?$