Show that, when $\displaystyle 2n$ terms of the series
$\displaystyle 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+...+(2n-1)^2-(2n)^2$
are grouped in pairs, an arithmetic series is formed.
Hello, SengNee!
Here's a rather primitive approach . . .
Show that, when $\displaystyle 2n$ terms of the series
$\displaystyle 1^2-2^2+3^2-4^2+...+(2r-1)^2-(2r)^2+ \cdots +(2n-1)^2-(2n)^2$
are grouped in pairs, an arithmetic series is formed.
Group in pairs: .$\displaystyle (1^2-2^2) + (3^2-4^2) + (5^2-6^2) + (7^2-8^2) + \cdots + \left([2n-1]^2 - [2n]^2\right) $
. . . . . . . . .$\displaystyle = \quad(-3)\quad+\quad(-7) \;\;\;+\;\;\; (-11) \;\;+\;\; (-15)\quad + \cdots +\quad \left(-[4n-1]\right) $
This is an arithmetic series with first term $\displaystyle a = \text{-}3$ and common difference $\displaystyle d = \text{-}4$
$\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(2k)^2$
$\displaystyle \sum_{k=1}^{2n}(-1)^{k+1}(k)^2$
I think the first one is false, the second one is correct.
Because for the first one, when $\displaystyle k=2n$,
$\displaystyle u_{2n}$
$\displaystyle =(-1)^{2n+1}(4n)^2$
$\displaystyle =-(4n)^2$
For the second one, when $\displaystyle k=2n$,
$\displaystyle u_{2n}$
$\displaystyle =(-1)^{2n+1}(2n)^2$
$\displaystyle =-(2n)^2$
Isn't it?
$\displaystyle S_{2n}=1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$
$\displaystyle S_{2n}=-(2n)^2+(2n-1)^2-(2n-2)^2+(2n-3)^2+...-2^2+1^2$
$\displaystyle 2S_{2n}=(1-4n^2)+(4n^2-4n-3)+(5+8n-4n^2)+$$\displaystyle (4n^2-12n-7)+...+(4n^2-4n-3)+(1-4n^2)$
$\displaystyle 2S_{2n}=(-4n-2)+(-4n-2)+...+(-4n-2)$
$\displaystyle 2S_{2n}=2n(-4n-2)$
$\displaystyle S_{2n}=n(-4n-2)$
$\displaystyle S_{2n}=-2n(2n+1)$
I do like this. Is it correct also?