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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

. . This is $\displaystyle S(k+1)$ . . . 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!