Proof by induction. how can I solve?

Question

How can I demonstrate this equality: $$1+2q+3q^2...+nq^{n-1}=\frac{1-(n+1)q^n+nq^{n+1}} {(1-q)^2} $$ My attempt: if $n=1$ $$1=\frac{1-2q+q^2} {(1-q)^2}=1$$ Now i demonstrate this equality: $$1+2q+3q^2...+nq^{n-1}+(n+1)q^n=\frac{1-(n+2)q^{n+1}+(n+1)q^{n+2}} {(1-q)^2}$$ $$1+2q+3q^2...+nq^{n-1}+(n+1)q^n=\frac{1-(n+1)q^n+nq^{n+1}} {(1-q)^2}+(n+1)q^n=\frac{1-(n+1)q^n+nq^{n+1}+(n+1)(1-q)^2q^n} {(1-q)^2}=\frac{1-nq^n+q^n+nq^{n+2}+nq^n+nq^{n+2}-2nq^{n+1}+q^n+q^{n+2}-2q^{n+1}} {(1-q)^2}=\frac{1-q^n(2n+2)-q^{n+1}(2n+2)+q^{n+2}(2n+1)} {(1-q)^2}=\frac {1-q^n(1-q)(2n+2)+q^{n+2}(2n+1)} {(1-q)^2} $$ Now I don't know to continue from here.

Answer 1

$$\begin{eqnarray*}(1-q)^2(1+2q+\ldots+nq^{n-1}) &=& (1-q)\left((1+2q+\ldots+nq^{n-1})-(q+2q^2+\ldots+n q^n\right)\\&=&(1-q)(1+q+q^2+\ldots+q^{n-1}-nq^n)\\&=&(1-q^n)-n(1-q)q^n\\&=&1-(n+1)q^n+nq^{n+1}\end{eqnarray*}$$ as wanted.

Answer 2

Given that $$1+2q+3q^2...+nq^{n-1}=\frac{1-(n+1)q^n+nq^{n+1}} {(1-q)^2}$$ you need to show that $$1+2q+3q^2...+(n+1)q^{n}=\frac{1-(n+2)q^{n+1}+(n+1)q^{n+2}} {(1-q)^2}$$ Note that you can substitute (1) into (2) to get $$(n+1)q^{n} = \frac{1-(n+2)q^{n+1}+(n+1)q^{n+2}} {(1-q)^2} - \frac{1-(n+1)q^n+nq^{n+1}} {(1-q)^2}$$ This can be simplified very quickly.

Answer 3

Expanding everything may not be good. You can do as the following :

$$1-(n+1)q^n+nq^{n+1}+(n+1)\color{red}{(1-q)^2q^n}$$ $$=1-(n+1)q^n+nq^{n+1}+(n+1)\color{red}{(1-2q+q^2)q^n}$$ $$=1-(n+1)q^n+nq^{n+1}+(n+1)\color{red}{(q^n-2q^{n+1}+q^{n+2})}$$ $$=1\color{blue}{-(n+1)q^n}+\color{green}{nq^{n+1}}+\color{blue}{(n+1)q^n}\color{green}{-2(n+1)q^{n+1}}+(n+1)q^{n+2}$$ $$=1\color{green}{-(n+2)q^{n+1}}+(n+1)q^{n+2}.$$