I will show that
\(2+5x+9x^2+14x^3+20x^4+...= \frac{1}{(1-x)^3} + \frac{1}{(1-x)^2} = \frac{2-x}{(1-x)^3} \).
Using the binomial series expansion
\((1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)
with \(n=-3\ and\ n=-2\) we get
\((1+x)^{-3}=1+(-3)x+\frac{-3(-4)}{2!}x^2+\frac{-3(-4)(-5)}{3!}x^3+\frac{-3(-4)(-5)(-6)}{4!}x^4+...\)
\(=1-3x+6x^2-10x^3+15x^4-...\)
and replacing \(x\ with\ -x\) in the above we will have
\(\frac{1}{(1-x)^3}=(1-x)^{-3}=1+3x+6x^2+10x+15x^4+...\)
And with \(n=-2, \)
\((1+x)^{-2}=1+(-2)x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+\frac{-2(-3)(-4)(-5)}{4!}x^4+...\)
\(=1-2x+3x^2-4x^3+5x^4-...\),
and replacing again x with - x, we get
\(\frac{1}{(1-x)^2}=(1-x)^{-2}=1+2x+3x^2+4x^3+5x^4+...\);
and finally adding the two series for n = -3 and n = -2, we obtain the function I gave above.
