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It's been a while since I have done a problem like this. I have this problem

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

So I have gotten this far:

Base Case: $$n=1$$

$$(-1)^2+1^2 = \frac{ (-1)^2 1(2) }{2}$$ $$ 1 = 1$$

Assume:

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

Prove: $n = n+1$

$$\sum_{i=0}^n+1 (-1)^ {i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I think that I can just replace the i's with (n+1) too

Which would be this

$$\sum_{i=0}^{n+1} (-1)^ {n+2} (n+1)^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I don't remember where to go from here.

4 Answers 4

4

You can most certainly not replace $i$s with $n$s. That's a major mistake. For example

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ while $$\sum_{i=1}^n n = n^2$$ and those are certainly not the same thing.

What you do have to do is replace $n$ with $n+1$ everywhere, including the sum, so what you have to prove is

$$\sum_{i=0}^{n+1} (-1)^{i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2}$$

while you can assume that

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

is true.

To prove the above equation, remember that

$$\sum_{i=0}^{n+1} a_i = \left[\sum_{i=0}^{n}a_i\right] + a_{n+1}$$

and now use this fact on your particular sum, then use the induction assumption that tells you something about the sum in parentheses.

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    So I should be getting $$\sum_{i=0}^n (-1)^{i+1} i^2 + (-1)^{n+2}(n+1)^2 $$ and then I would replace the summation with the original formula correct? – 2017-02-07
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    @kingcobra1986 Yeah, there you go. It should be simple from there on. – 2017-02-07
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    So then I need to solve $$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} + (-1)^{n+2}(n+1)^2 $$ Where the right side should essentially reduce down to the RHS of the original equation. – 2017-02-07
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    @kingcobra1986 I don't know how you got to that particular equation. What you need to prove is$$\sum_{i=0}^{n+1} (-1)^{i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2}$$ and you already know what the left side is eual to. Now use the induction hypothesis. – 2017-02-07
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    So like this:$$\frac{(-1)^{n+1}n(n+1)}{2} + (-1)^{n+2}(n+1)^2 = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$$ – 2017-02-07
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    @kingcobra1986 Yes, that's what you need to prove. It's simple algebrait manipulation from here on, maybe just split the cases when $n$ is even and odd... – 2017-02-07
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You have shown it is true for the base case. Now, let's move on to the inductive step: Note that you should be using $n=k$ and $n=k+1$ instead of $n=n$ and $n=n+1$.

Assume true for $n=k$: $$\sum_{i=0}^{k} (-1)^{i+1} i^2=\frac{(-1)^{k+1} k(k+1)}{2} \tag{1}$$ For $n=k+1$: $$\sum_{i=0}^{k+1} (-1)^{i+1} i^2=\frac{(-1)^{k+2} (k+1)(k+2)}{2} \tag{2}$$ Now, add the $k+1$ term of the series to equation $(1)$: $$\sum_{i=0}^{k} (-1)^{i+1} i^2+\color{blue}{(-1)^{k+2}(k+1)^2}=\frac{(-1)^{k+1} k(k+1)}{2}+\color{blue}{(-1)^{k+2}(k+1)^2}$$ $$\Rightarrow \sum_{i=0}^{k+1} (-1)^{i+1} i^2=\frac{(-1)^{k+1} k(k+1)}{2}+\color{blue}{(-1)^{k+2}(k+1)^2}\tag{3}$$ Now, try to make the RHS of equation $(3)$ equal to equation $(2)$.

Can you continue?

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If $n$ is even, $$\begin{align} \sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2+\cdots+(n-1)^2-n^2\\ &=(1-2)(1+2)+(3-4)(3+4)\cdots+((n-1)-n)((n-1)+n)\\\\ &=-(1+2+3+4+\cdots+(n-1)+n)\\ &=-\frac{n(n+1)}2\end{align}$$

If $n$ is odd, $$\begin{align} \sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2-\cdots-(n-1)^2+n^2\\ &=(-0+1)(0+1)+(-2+3)(2+3)+\cdots+(-(n-1)+n)((n-1)+n)\\\\ &=1+2+3+4+\cdots+(n-1)+n\\ &=\frac{n(n+1)}2\end{align}$$

Hence $$\sum_{i=0}^n (-1)^{i+1}i^2=(-1)^{n+1}\frac {n(n+1)}2$$

-1

See the summation of the n+1 terms is just the summation of n terms (which is known) + $(-1)^{n+2}(n+1)$. Now you see the minus sign $(-1)^{n+1}$can be taken common the remaining term is :

$ \frac{n(n+1)}{2} -(n+1)^2 = (n+1) .\frac{n-2(n+1)}{2} = -\frac{(n+1) (n+2)}{2} $.

Thus at last the minus sign turns out to e $(-1)^{n+2}$ and hence proved.