3

I'd like to find a closed-form formula for $$\sum_{k=1}^{n-2} [(-k + (-1)^k)^2 - 3]$$ using the indefinite summation method. That would be fine besides this annoying factor $(-1)^k$ for the expanded form of the summands: $k^2 - 2(-1)^kk + 1 - 3$. I am fine with using antidifferences to $k^1, k^2$ but here I am not sure.

Is there some neat way to find the antidifference of this sequence?

Robert Barg
  • 281
  • 1
  • 7

1 Answers1

1

We have \begin{align*} \Delta\left[(-1)^kk\right]&=(-1)^{k+1}(k+1)-(-1)^kk\\ &=-2(-1)^kk-(-1)^k. \end{align*} Rearranging, we find \begin{align*} (-1)^kk&=-\frac{1}{2}(-1)^k-\frac{1}{2}\Delta\left[(-1)^kk\right]\\ &=\Delta\left[\frac{1}{4}(-1)^k\right]-\Delta\left[\frac{1}{2}(-1)^kk\right]\\ &=\Delta\left[\frac{1}{4}(-1)^k-\frac{1}{2}(-1)^kk\right]\\ &=\Delta\left[\frac{(-1)^k}{4}(1-2k)\right]. \end{align*} Hence \begin{align*} \sum_{k=1}^{n-2}\left[(-k+(-1)^k)^2-3\right]&=\sum_{k=1}^{n-2}\left[k^2-2(-1)^kk-2\right]\\ &=\left[\frac{k(k-1)(2k-1)}{6}-2\left(\frac{(-1)^k}{4}(1-2k)\right)-2k\right]_{k=1}^{k={n-1}}\\ &=\left(\frac{(n-2)(n-1)(2n-3)}{6}-\frac{(-1)^n}{2}(2n-3)-2n+2\right)-\left(-\frac{5}{2}\right)\\ &=\frac{(2n-7)(n^2-n-3)}{6}-\frac{(-1)^n}{2}(2n-3).\\ \end{align*}

Freddie
  • 371