# Math Help - Series Problem

1. ## Series Problem

Find:
1 + 2a + 3a² + 4a³ + ... + na^(n-1)

2. Hello, bearej50!

Find: . $1 + 2a + 3a^2 + 4a^3 + \hdots + na^{n-1}$

. . . . $\text{We have: }\quad S \;=\; 1+2a + 3a^2 + 4a^3 + \hdots + na^{n-1}$

$\text{Multiply by }a\!:\;\;\;aS \;=\;\qquad a + 2a^2 + 3a^3 + \hdots + (n-1)a^n + na^{n+1}$

$\text{Subtract: }\;S - aS \;=\;\underbrace{1 + a + a^2+ a^3 + a^n}_{\text{geometric series}} - na^{n+1}$

. . . $(1-a)S \;=\;\frac{1-a^n}{1-a} - na^{n+1} \;=\;\frac{1-a^n - na^{n+1}(1-a)}{1-a}$

$\text{Therefore: }\;S \;=\;\frac{1 - a^n - na^{n+1}(1-a)}{(1-a)^2}$

3. Let $f(a)=1+2a + 3a^2 + 4a^3 + \hdots + na^{n-1}$. Integrate both sides, this is $F(a)=a+a^2+ \hdots a^n=\frac{1-a^{n+1}}{1-a}-1$. Differentiating yields the result
$f(a)=\frac{1-a^{n+1}}{(1-a)^2}-\frac{(n+1)a^n}{1-a}=\frac{1-a^n-na^n+na^{n+1}}{(1-a)^2}$ and this is your answer.