Give a Sequence, Find Its Generating Function: $\binom{8}{1},2\binom{8}{2},3\binom{8}{3},\dots , 8\binom{8}{8}$

Question

The Full Question

Find the generating function(closed form) of the following sequence:

$\binom{8}{1},2\binom{8}{2},3\binom{8}{3},\dots , 8\binom{8}{8}$

My Work

The open form of this generating function is:

$1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$

This can be re-written as:

$0\binom{8}{0}x^0+1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$

Which in Sigma Notation is:

$\sum_{k=0}^{8}k\binom{8}{k}x^k$

Part of this series is obviously binomial, which has a closed form of $(1+x)^8$ I don't know how to deal with the $k$ that is always being multiplied in. I thought I could maybe use the multiplication rule for sums, but it doesn't seem to apply here because we have $a_1b_1$ instead of $a_1b_n$

My Problem

How do we deal with the k that is being multiplied in there constantly? How can I get a nice closed form generating function with that $k$ getting in my way?

Answer 1

Hint: Say the generating function is $ f(x)$. What do you know about $ \int \frac{f(x) - 8}{x} \, dx $?

Answer 2

Hint: $i { 8 \choose i } = 8 { 7 \choose i-1 } $

You should be able to quickly conclude that the generating function is

$ 8 ( 1 + x) ^ 7 x $