# Transforming the Binomial Theorem into $\sum_{k=1}^n k3^k {n \choose k}$

By setting $x$ equal to the appropriate values in the binomial expansion (or one of its derivatives, etc.), evaluate

$$\sum_{k=1}^n k3^k {n \choose k}$$

Basically this problem is tasking me with taking the Binomial Theorem:

$$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$

And performing manipulations until it reaches the given form, in which case a value of $x$ can be determined.

By taking the derivative, we have that:

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

By changing the bounds:

$$n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)x^{(k-1)-1}$$

$$n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)\frac{x^k}{x^2}$$

This could also become:

$$x^2n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)x^k$$

Which is pretty close to the desired form, but I'm not really sure what to do with the $k-1$ term inside ${n \choose k-1}$. I can further simplify by doing:

$$x^2n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} kx^k - x^k$$

And I'm not sure if this is a valid simplification, given the Sigma:

$$x^2n(1+x)^{n-1} + x^k = \sum_{k=1}^n {n \choose k-1} kx^k$$

How can I achieve this desired form? Thanks so much! I appreciate the help!

• I think multiply by $x$ the both sides of $$n(1+x)^{n-1} = \sum_{k=0}^n {n \choose k} kx^{k-1}$$ would work. – Cave Johnson Sep 20 '16 at 5:34
• A combinatorial way to get this through is by using $k\binom nk = n \binom{n-1}{k-1}$. – Sungjin Kim Sep 20 '16 at 5:58

Evaluating the expression at $x=3$ gives the result.
Consider $$\sum_{k=1}^n k {n \choose k}x^k=x\sum_{k=1}^n k {n \choose k}x^{k-1}=x \left(\sum_{k=1}^n {n \choose k}x^{k} \right)'=x\left((1+x)^n-1\right))'=nx(1+x)^{n-1}$$