Proving (1+x)g'(x) = kg(x) using Binomial Series | Homework Solution

[toothpaste666]

Homework Statement

$g(x) = \sum_{n=0}^\infty \binom{k}{n} x^n$

$g'(x) = k\sum_{n=0}^\infty \binom{k-1}{n} x^n$

prove that (1+x)g'(x) = kg(x)

The Attempt at a Solution

$k(1+x)\sum_{n=0}^\infty \binom{k-1}{n} x^n$

distribute

$k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + x\sum_{n=0}^\infty \binom{k-1}{n} x^n]$

$k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + \sum_{n=0}^\infty \binom{k-1}{n} x^{n+1}]$

$k[\sum_{n=0}^\infty \binom{k-1}{n} x^n + \sum_{n=1}^\infty \binom{k-1}{n-1} x^n]$

This is where I am stuck. I want to be able to pull out the x^n and add $\binom{k-1}{n} + \binom {k-1}{n-1}$ because I already know when you add those it gives you $\binom{k}{n}$ and I would have my answer, but one series starts at one and the other 0 so I can't pull them out. Please help

 

[vela]

Just separate out the first term:
$$\sum_{n=0}^\infty \begin{pmatrix}k-1 \\ n\end{pmatrix}x^n = 1 + \sum_{n=1}^\infty \begin{pmatrix}k-1 \\ n\end{pmatrix}x^n$$

 

[toothpaste666]

Ahhh okay thank you! So:

$k[1 + \sum_{n=1}^\infty \binom{k-1}{n} x^n + \sum_{n=1}^\infty \binom{k-1}{n-1} x^n]$

$k[1 + \sum_{n=1}^\infty x^n ( \binom{k-1}{n} + \binom{k-1}{n-1})]$


$k[1 + \sum_{n=1}^\infty x^n \binom{k}{n}]$


$k\sum_{n=0}^\infty x^n \binom{k}{n}$