# Prove that $1+2q+3q^2+…+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}$

Prove:

$$1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}$$

Hypothesis:

$$F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}$$

Proof:

$$P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2}$$ $$P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2}$$ $$P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} |$$

Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next.

$$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2}$$
$$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2}$$

I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.

You can also prove by induction. Assume true for $n$, and show for $n+1$.

$$1+2q+3q^2+\cdots+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}.$$

Then,

$$1+2q+3q^2+\cdots+nq^{n-1}+(n+1)q^n = \frac{1-(n+1)q^n+nq^{n+1}+(n+1)q^n(1-q)^2}{(1-q)^2}= \frac{1-(n+2)q^{n+1}+(n+1)q^{n+2}}{(1-q)^2}.$$

Thus the result holds for $n+1$. Since the result holds for $n=1$, the proof is complete.

Hint: $1+2q+3q^2+\ldots+nq^{n-1}= (q+q^2+\ldots+q^{n})'$

\begin{align} S&=1+2q+3q^2+\qquad\cdots\qquad \qquad+nq^{n-1}\\ qS&=\qquad q+2q^2+3q^3+\cdots +\quad(n-1)q^{n-1}+nq^n \\ \text{Subtracting,}&\\ (1-q)S&=1+\;\ q \ +\ q^2 +\ q^3+\cdots \qquad \qquad +q^{n-1}-nq^n\\ &=\frac {\;\ 1-q^n}{1-q}-nq^n\\ S&=\frac{1-q^n-nq^n(1-q)}{(1-q)^2}\\ &=\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}\qquad\blacksquare\end{align}

Here is an alternative approach, motivated by the fact that it's usually wise in such problems to multiply the $(1-q)$ factors through:

\begin{align} (1-q)^2(1+2q+3q^2+\cdots +nq^{n-1}) &=(1-q)\cdot (1-q)(1+2q+3q^2+\cdots +nq^{n-1})\\ &=(1-q)\cdot (1+q+q^2+\cdots +q^{n-1}-nq^{n})\\ &=1-(n+1)q^{n}+nq^{n+1}. \end{align} Induction can be used to verify that multiplying by $(1-q)$ cancels like terms as above.

Hint: Consider the finite geometric series given by

$$Q(q,p)=\sum_{k=1}^p q^k=\frac{1-q^{p+1}}{1-q}$$

Consider $\frac{d}{dq}G(q,n)$ and see where it takes you...

Ok, so, I had some mistakes in my first approach to the problem, specially with signs, but I already proved it, so, here it is.

Hypothesis :

$$F(x) = 1+2q+3q^2+...+xq^{x+1}=\frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}$$

Proof:

$P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2}+(x+1)q^x = \frac{1-(x+2)q^{x+1}+(x+1)q^{x+2}}{(1-q)^2}$

$P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q^2)]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+(x+1)q^{x+2}}{(1-q)^2}$

$P3 | \frac{xq^{x+1}+[-(x+1)+(x+1)(1^2-\binom{2}{1}*1*q+q^2)]q^x+1}{(1-q)^2} = \frac{(x+1)q^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2}$

$P4 | \frac{x\color{red}{q^{x+1}}+[xq^2-2xq+q^2-2q]\color{red}{q^x}+1}{(1-q^2)} = \frac{(x+1)\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2}$

$P5 | \frac{xq^{x+1} + xq^{x+2} - 2xq^{x+1} + q^{x+2} - 2q^{x+1}+1}{(1-q)^2} = \frac{xq^{x+2}+q^{x+2}-(xq^{x+1}+2q^{x+1})+1}{(1-q)^2}$

$∴ | \frac{xq^{x+2}-xq^{x+1}+q^{x+2}-2q^{x+1}+1}{(1-q)^2} = \frac{xq^{x+2}-xq^{x+1}+q^{x+2}-2q^{x+1}+1}{(1-q)^2}$

Quod erat demonstrandum