1. ## Induction with summing

Now i usually can do induction problems but i cant figure out what to do when theres summing involved. Can anyone help with the following question?

Prove by induction:
$\sum_{k=1}^n (-1)^2 k^2 = \frac{1}{2}(-1)^nn(n+1)$

It works for n = 1 so we have our inductive phase...
Now assuming n is true...
How do we prove this for n + 1?

Now i usually can do induction problems but i cant figure out what to do when theres summing involved. Can anyone help with the following question?

Prove by induction:
$\sum_{k=1}^n (-1)^2 k^2 = \frac{1}{2}(-1)^nn(n+1)$

It works for n = 1 so we have our inductive phase...
Now assuming n is true...
How do we prove this for n + 1?
you're missing something. (-1)^2 = 1, it would make no sense to write it. so there should probably be an n in the power or something. so you left that off, maybe other things

3. It is $
\sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
{2}
$
, right?

Suppose $
\sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
{2}
$
holds for n (1)

We'll now show that: $
\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{\left( {n + 2} \right) \cdot \left( {n + 1} \right)}}
{2}
$
is also true

Indeed: $
\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } + \left( { - 1} \right)^{n + 1} \cdot \left( {n + 1} \right)^2
$

But, by (1) we have: $
\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
{2} + \left( { - 1} \right)^{n + 1} \cdot \left( {n + 1} \right)^2
$

So: $
\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \left( {n + 1} \right) \cdot \left[ {\frac{n}
{2} - \left( {n + 1} \right)} \right] = \left( { - 1} \right)^n \left( {n + 1} \right) \cdot \left[ {\frac{{ - n - 2}}
{2}} \right]

$

$
\sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^{n + 1} \cdot \frac{{\left( {n + 1} \right) \cdot \left( {n + 2} \right)}}
{2}
$

Prove by induction: . $\sum_{k=1}^n (-1)^2 k^2 \:= \:\frac{1}{2}(\text{-}1)^nn(n+1)$
$S(n)\!:\;\;-1^2 + 2^3 - 3^2 + 4^2 + \cdots + (\text{-}1)^nn^2 \;=\;\frac{1}{2}(\text{-}1)^nn(n+1)$

Verify $S(1)\!:\;\;-1^2 \:=\:\frac{1}{2}(\text{-}1)(1)(2)\quad\Rightarrow\quad -1 \:=\:-1$ . . . True!

Assume $S(k)\!:\;\;-1^2 + 2^2 - 3^2 + 4^2 + \cdots + (\text{-}1)^kk^2 \;=\;\frac{1}{2}(\text{-}1)^kk(k+1)$

Add $(\text{-}1)^{k+1}(k+1)^2$ to both sides:

. . $-1^2 + 2^3 - 3^2 + 4^2 + \cdots + (\text{-}1)^{k+1}(k+1)^2 \;=\;\frac{1}{2}(\text{-}1)^kk(k+1) + (\text{-}1)^{k+1}(k+1)^2$ . [1]

Factor the right side: . $\frac{1}{2}(\text{-}1)^k(k+1)\,\bigg[k - 2(k+1)\bigg] \;=\;\frac{1}{2}(\text{-}1)^k(k+1)[-k-2]$

. . $=\;\frac{1}{2}(\text{-}1)^k(k+1)(\text{-}1)[k+2] \;=\;\frac{1}{2}(\text{-}1)^{k+1}(k+1)(k+2)$

Hence, [1] becomes: . $-1^2 + 2^3 - 3^2 + 4^3 + \cdots + (\text{-}1)^{k+1}(k+1)^2 \;=\;\frac{1}{2}(\text{-}1)^{k+1}(k+1)(k+2)$

. . This is $S(k+1)$ . . . The inductive proof is complete.

5. Thanks everybody, the (-1)^2 was a typo, was meant to be, as you all guessed, (-1)^k. Cheers!