differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3. Thanks
$\displaystyle \frac{1}{1-x} = \sum^{\infty}_{n=0} x^n \text{ for }|x| < 1$
If you differentiate $\displaystyle \frac{1}{1-x}$ twice you get $\displaystyle \frac{2}{(1-x)^3}$
You just differentiate each term in the sum above and divide by 2. After a few terms you should see a pattern then just turn it into an infinite sum.
$\displaystyle \frac{1}{(1-x)} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 +$ etc...
So differentiate both sides...
$\displaystyle -\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5 x^4 + 6x^5 +$ etc...
differentiate again...
$\displaystyle \frac{2}{(1-x)^3} = 2 + 6x + 12x^2 + 20x^3 + 30x^4 +$ etc...
This is a series of the form...
$\displaystyle \sum_{n=0}^{\infty} (n+1)(n+2)x^n$.
How did I get this? Just differentiate $\displaystyle x^n$ twice. You don't even need to all that stuff I did above. Just it's good to see it done so you know what's going on.
Differentiating $\displaystyle x^n$ twice gives you $\displaystyle n(n-1)x^{n-2}$. I.e. $\displaystyle (n+2)(n+1)x^n$
So summarizing...
$\displaystyle \frac{2}{(1-x)^3} = \sum_{n=0}^{\infty} (n+2)(n+1)x^n $
=> $\displaystyle \frac{}{(1-x)^3} = \frac{1}{2} \sum_{n=0}^{\infty} (n+2)(n+1)x^n$