Find the formula for the sum $1^2 - 2^2 + 3^2 - 4^2 + \dots + (-1)^{n-1}n^2$

Question

I was given an assignment of finding the sum of $$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \dots + (-1)^{n-1}n^2 $$ Where $n$ is an odd number.

I've already checked the answer in wolframalpha which is $$\sum_{k=1}^n (-1)^{k-1}k^2 = -\frac{(-1)^n n(n+1)}{2}$$

and was wondering if anyone would be willing to show me how to solve the exercise step-by-step.

I've tried grouping positive and negative terms and got pretty close to the actual answer but then ran out of ideas $$\sum_{k=1}^{n/2} (2n-1)^2-(2n)^2 = -\frac{n(n+1)}{2}$$

Thank you all.

Answer 1

Snake oil: \begin{align} \sum_{n=1}^\infty \color{red}{\sum_{k=1}^n (-1)^{k-1} k^2} z^n &= \sum_{k=1}^\infty (-1)^{k-1} k^2 \sum_{n=k}^\infty z^n \\ &= \sum_{k=1}^\infty (-1)^{k-1} k^2 \frac{z^k}{1-z} \\ &= \frac{-1}{1-z}\sum_{k=1}^\infty k^2 (-z)^k \\ &= \frac{-1}{1-z}\cdot\frac{-z(1-z)}{(1+z)^3} \\ &= \frac{z}{(1+z)^3} \\ &= z\sum_{n=0}^\infty \binom{n+2}{2}(-z)^n \\ &= -\sum_{n=0}^\infty \binom{n+2}{2}(-z)^{n+1} \\ &= \sum_{n=1}^\infty \color{red}{-(-1)^n\binom{n+1}{2}}z^n \end{align}

Answer 2

Your answer assumes that $n$ is even. However, when $n$ is odd, the answer becomes $$n^2 + \sum_{k=1}^{\lfloor n/2 \rfloor} (2k-1)^2-(2k)^2 = -\frac{n(n-1)}{2}+n^2 = \frac{n^2+n}{2} = \frac{(n+1)n}{2}$$

The answer is $-\frac{n(n+1)}{2}$ when $n$ is even, and $\frac{(n+1)n}{2}$ when $n$ is odd, which makes the answer: $$-\frac{(-1)^n(n+1)n}{2}$$

Answer 3

Say, for even number of terms we split the series as

$$\begin{align*} S_{\text{pair}} = 1^2 -2^2 + 3^2-4^2......\text{with even number of terms} \end{align*}$$

+--------+--------------+---------+
|        |              |-1*Below |
+--------+--------------+---------+
|1^2- 2^2|  (1-2)(1+2)  |  3      |
+--------+--------------+---------+
|3^2- 4^2|  (3-4)(3+4)  |  7      |
+--------+--------------+---------+
|5^2- 6^2|  (5-6)(5+6)  |  11     |
+--------+--------------+---------+

$$\begin{align*} S_{\text{pairs}} & = 1^2 -2^2 + 3^2-4^2......\text{with even number of terms}\\ & =(-4)\times\sum_{k = 1}^{\frac n2}(2k- 1) \\ & = -\frac {n(n+1)}{2}\\ \end{align*}$$

$$\begin{align*} \color{blue}{S_{\text{unpaired with n-terms}}} &= \color{blue}{S_{\text{paired with (n-1) terms}} + n^2}\\ & = -\frac {(n-1)(n-1+1)}{2} + n^2\\ & = \frac {n(n+1)}2\\ \end{align*}$$ Combining both $$S= -(-1)^n\times \frac {n(n+1)}2$$

Answer 4

We have :\begin{aligned}\sum_{k=0}^{n}{\left(-1\right)^{k-1}k^{2}}&=\sum_{k=0}^{n-1}{\left(-1\right)^{k}\left(k+1\right)^{2}}\\ \sum_{k=0}^{n}{\left(-1\right)^{k-1}k^{2}}&=-\sum_{k=0}^{n}{\left(-1\right)^{k-1}k^{2}}-2\sum_{k=0}^{n}{\left(-1\right)^{k-1}k}+\sum_{k=0}^{n}{\left(-1\right)^{k}}-\left(-1\right)^{n}\left(n+1\right)^{2}\\ \iff 2\sum_{k=0}^{n}{\left(-1\right)^{k-1}k^{2}}&=-2\sum_{k=0}^{n}{\left(-1\right)^{k-1}k}+\frac{1+\left(-1\right)^{n}}{2}-\left(-1\right)^{n}\left(n+1\right)^{2}\end{aligned}

Let's find a closed form for $ \sum\limits_{k=0}^{n}{\left(-1\right)^{k-1}k} $, using the same trick :

\begin{aligned}\sum_{k=0}^{n}{\left(-1\right)^{k-1}k}&=\sum_{k=0}^{n-1}{\left(-1\right)^{k}\left(k+1\right)}\\ \sum_{k=0}^{n}{\left(-1\right)^{k-1}k}&=-\sum_{k=0}^{n}{\left(-1\right)^{k-1}k}+\sum_{k=0}^{n}{\left(-1\right)^{k}}-\left(-1\right)^{n}\left(n+1\right)\\ \iff 2\sum_{k=0}^{n}{\left(-1\right)^{k-1}k}&=\frac{1+\left(-1\right)^{n}}{2}-\left(-1\right)^{n}\left(n+1\right)\\ \iff \ \ \sum_{k=0}^{n}{\left(-1\right)^{k-1}k}&=\frac{1-\left(-1\right)^{n}\left(2n+1\right)}{4}\end{aligned}

Thus : $$ 2\sum_{k=0}^{n}{\left(-1\right)^{k-1}k^{2}}=\frac{\left(-1\right)^{n}\left(2n+1\right)-1}{2}+\frac{1+\left(-1\right)^{n}}{2}-\left(-1\right)^{n}\left(n+1\right)^{2}=\left(-1\right)^{n+1}\left(n+1\right) $$

Answer 5

It doesn't seem right to post a full answer, but as so many answers have already been posted, and as this question will soon be closed as a duplicate, anyway, I hope I may be forgiven for adding to the carnage.

Denote the sum by $S(n),$ regardless of whether $n$ is odd or even.

For every integer $r,$ $$ (r + 1)^2 - r^2 = r + (r + 1). $$

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

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