Pull constant out of summation

Question

In the solution of my homework there's this step that I don't understand:

$$\begin{align}\mathsf {Var}[X] & = \sum_{k=1}^{n}\dfrac{n(k-1)}{(n-k+1)^2}\\ &= n \sum_{k=1}^{n}\dfrac{(n-k)}{k^2}\\&= n^2 \sum_{k=1}^{n}\dfrac{1}{k^2}-n \sum_{k=1}^{^n}\dfrac{1}{k}\end{align}$$

I would have thought, that when I pull $n$ out of the $\sum$ that I'd just get: $n\sum_{k=1}^n\frac{(k-1)}{(n-k+1)^2}$.

Also I don't understand the second line. Could someone please tell me what steps are missing and how they solved this? Thanks!

Answer 1

They've just preformed the distribution and substitution at the same time rather than two steps.

$$\begin{align}\mathsf {Var}[X] & = \sum_{k=1}^{~~n}\dfrac{n(k-1)}{(n-k+1)^2} &&\text{as given} \\ &= n\sum_{k=1}^n\dfrac{(k-1)}{(n-k+1)^2} &&\text{distribution} \\ &= n \sum_{k=1}^{n}\dfrac{(n-k)}{k^2} && \text{substitution: } k\gets n-k+1 \\&= n \sum_{k=1}^{n}\dfrac{n}{k^2}-n \sum_{k=1}^{n}\dfrac{k}{k^2}&&\text{commutation and association}\\&= n^2 \sum_{k=1}^{n}\dfrac{1}{k^2}-n \sum_{k=1}^{n}\dfrac{1}{k}&&\text{distribution and rationalisation}\end{align}$$

Answer 2

Alternatively, note that the summation is reversed: $$\begin{align}\color{red}{\mathsf {Var}[X]} & \color{red}{= \sum_{k=1}^{n}\dfrac{n(k-1)}{(n-k+1)^2}}\\ &=n\left(\frac{1-1}{n^2}+\frac{2-1}{(n-1)^2}+\frac{3-1}{(n-2)^2}+\cdots+\frac{n-2-1}{3^2}+\frac{n-1-1}{2^2}+\frac{n-1}{1^2}\right) \\ &=n\left(\frac{n-1}{1^2}+\frac{n-2}{2^2}+\frac{n-3}{3^2}+\cdots+\frac{2}{(n-2)^2}+\frac{1}{(n-1)^2}+\frac{0}{n^2}\right) \\ &\color{red}{=n \sum_{k=1}^{n}\dfrac{(n-k)}{k^2}}\\ &=n \sum_{k=1}^{n}\left(\dfrac{n}{k^2}-\frac{k}{k^2}\right)\\ &\color{red}{= n^2 \sum_{k=1}^{n}\dfrac{1}{k^2}-n \sum_{k=1}^{^n}\dfrac{1}{k}}.\end{align}$$