Proof alternating sum of squares is alternating sign of sum

Question

I'm trying to prove by induction that $1-4+9-...\pm n^2 = \pm(1+2+...+n)$. The base-case is obvious, and the formula that I write this as is

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

If we assume this holds up to $n$ then I try to prove the equation true for $n+1$ but get stuck. I have

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

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

At this point I can't think of anyth ing that sounds like a good idea. I suppose I could try to use what I have to get a formula for $n+1$ like moving everything away from this in the inductive hypothesis, $n+1=\pm(1-4+...\pm n^{2})-2-3-...-(n-1)$ but that sounds like I'll just be heading in a circle.

For guidance I can try setting this equal to my goal and reduce it to a known equation, but that doesn't seem to help.

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

iff

$$n^{2}+2n+1 = n+1+2\sum_{i=1}^{n}i$$

iff

$$n^{2} = n+ 2\sum_{i=1}^{n-1}i$$

but this is not obviously true and I can't see where to go from here either.

Answer 1

Let's assume $1^2 - 2^2 + 3^2 - \ldots + (n-1)^2 = 1 + 2 + 3 + \ldots + (n - 1)$, and see what happens when we subtract $n^2$. Note that subtraction must come next, as we're assuming we have a positive sum.

To use this proof, you must be able to use the fact that $1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}$.

Subtracting $n^2$, we get $$\frac{n(n - 1)}{2} - n^2 = \frac{n^2 - n}{2} - \frac{2n^2}{2} = \frac{-n^2 - n}{2} = -\frac{n(n+1)}{2} = -(1 + 2 + \ldots + n).$$

Does that help?

Answer 2

Recall that $$\sum_{k = 1}^{n} k = \frac{n(n + 1)}{2}$$

For the induction step: Let the induction hypothesis be $$\sum_{k = 1}^{m} (-1)^{k + 1} k^2 = (-1)^{m + 1}\sum_{k = 1}^{m} k$$ for some $m \in \mathbb{N}$. Let $n = m + 1$. Then \begin{align*} \sum_{k = 1}^{m + 1} (-1)^{k + 1} k^2 & = \sum_{k = 1}^{m} (-1)^{k + 1} k^2 + (-1)^{m + 2} (m + 1)^2\\ & = (-1)^{m + 1}\sum_{k = 1}^{m} k + (-1)^{m + 2} (m + 1)^2 && \text{induction hypothesis}\\ & = (-1)^{m + 1} \frac{m(m + 1)}{2} + (-1)^{m + 2} (m + 1)^2\\ & = (-1)^{m + 2}(m + 1)\left[-\frac{m}{2} + m + 1\right]\\ & = (-1)^{m + 2}\frac{m + 1}{2}(-m + 2m + 2)\\ & = (-1)^{m + 2}\frac{(m + 1)(m + 2)}{2}\\ & = (-1)^{m + 2}\sum_{k = 1}^{m + 1} k \end{align*}

Answer 3

A short answer to your question:

Note that $-r^2+(r+1)^2-r^2=r+(r+1)$.

Hence, for odd $n$,

$$\begin{align} &1^2 \underbrace{-2^2+3^2}_{2+3} \underbrace{-4^2+5^2}_{4+5}+\cdots+\underbrace{-(n-1)^2+n^2}_{(n-1)+n} &=1+2+3+4+\cdots+n\\ \end{align}$$

and for even $n$,

$$\begin{align} &\underbrace{1^2-2^2}_{-1-2}+ \underbrace{3^2-4^2}_{-3-4}+\cdots+\underbrace{(n-1)^2-n^2}_{-(n-1)-n} &=-\bigg(1+2+3+4+\cdots+n\bigg) \end{align}$$ .

In summary, $$1^2-2^2+3^3-4^4+\cdots+(-1)^{n-1} n^2 =(-1)^{n-1}\bigg(1+2+3+4+\cdots+n\bigg)\;\blacksquare$$ or $$\sum_{r=1}^n (-1)^{r-1}r^2=(-1)^{n-1}\sum_{r=1}^n r=(-1)^{n-1}\binom {n+1}2$$