Use the principle of Mathematical Induction to show: $$\sum _ { k = 1 } ^ { n } k x ^ { k - 1 } = \frac { 1 - ( n + 1 ) x ^ { n } + n x ^ { n + 1 } } { ( 1 - x ) ^ { 2 } }$$ for every $n\in\mathbb{N}$
I understand the concept of induction in general, but I'm not sure how to use induction in this scenario
Hint. First, note that the statement is trivially true for $n=0$ under the interpretation $\sum_{k=1}^{0}f(k)=0$ (and $0^0 = 1$ if $x=0$). This the base case. As for the inductive case, suppose the statement holds for some particular positive integer $n$. Then, $$ \sum_{k=1}^{n+1}kx^{k-1} =\left(\sum_{k=1}^{n}kx^{k-1}\right)+\left(n+1\right)x^{n} =\frac{1-\left(n+1\right)x^{n}+nx^{n+1}}{\left(1-x\right)^{2}}+\left(n+1\right)x^{n}. $$ I leave it to you to massage the rightmost expression in the above equality into the form $$ \frac{1-\left(n+2\right)x^{n+1}+\left(n+1\right)x^{n+2}}{\left(1-x\right)^{2}}, $$ establishing the inductive case. Now, conclude by the principle of induction.
It's good practice to point out that the claim is meaningless at $x=1$.