How do you find the general expression of a derived sum?

Question

We got an exercice about maximum and minimum and there is a question about a sum. The goal is to find the general expression of the sum ( without the sigma ). And there is a kx^k form. We need to derive and I can understand why but I just can't find how you derive a sum.

Answer 1

First, the sum of geometric progression is $$ \sum_{k=1}^{n}x^k = \dfrac{x(1-x^n)}{1-x}.\tag{1} $$ Then we can rewrite the sum $\displaystyle\sum_{k=1}^{n} kx^k$ in the form $$ \sum_{k=1}^{n} kx^k = \sum_{k=1}^{n} x^k + \sum_{k=2}^{n} x^k + \sum_{k=3}^{n} x^k + \ldots + \sum_{k=n-1}^{n} x^k + (x^n). \tag{2} $$ Based on $(1)$ we continue: $$ \sum_{k=1}^{n} kx^k = \dfrac{x(1-x^n)}{1-x}+ \dfrac{x^2(1-x^{n-1})}{1-x}+\dfrac{x^3(1-x^{n-2})}{1-x}+ \ldots + \dfrac{x^{n-1}(1-x^2)}{1-x}+ \dfrac{x^n(1-x)}{1-x} $$ $$ = \dfrac{x-x^{n+1}}{1-x}+ \dfrac{x^2-x^{n+1}}{1-x}+\dfrac{x^3-x^{n+1}}{1-x}+ \ldots + \dfrac{x^{n-1}-x^{n+1}}{1-x}+ \dfrac{x^{n}-x^{n+1}}{1-x} $$ $$ = \dfrac{1}{1-x}\left( (x-x^{n+1}) + (x^2-x^{n+1}) + (x^3-x^{n+1}) + \ldots + (x^{n-1}-x^{n+1}) + (x^n - x^{n+1})\right) $$ $$ = \dfrac{1}{1-x}\left( \sum_{k=1}^{n}x^k-nx^{n+1}\right) $$ $$ = \dfrac{1}{1-x}\left( \dfrac{x(1-x^n)}{1-x}-nx^{n+1}\right).\tag{3} $$ And many equivalent modifications: $$ \dfrac{1}{x-1}\left(nx^{n+1} -\dfrac{x(x^n-1)}{x-1}\right); $$ $$ \dfrac{x}{x-1}\left(nx^n -\dfrac{(x^n-1)}{x-1}\right); $$ $$ \dfrac{x}{(x-1)^2}\left(nx^{n+1} - nx^n -(x^n-1)\right); $$ $$ \dfrac{x}{(x-1)^2}\left(nx^{n+1} - (n+1)x^n +1\right); $$ (see WolframAlpha link to check the formula).