Prove the following combinatorial identity by differentiating the Binomial Theorem

Question

$$(2^{n-1} + 1)\sum_{k=0}^{n-2}2^k = \frac{1}{n}\sum_{k=2}^n k \binom{n}{k}3^{k-1} $$

I've tried differentiating the Binomial Theorem and substituting $x = 3, y=1$ and got the RHS. But can't seem to get this to equal the LHS.

I'm stuck. Any help would be greatly appreciated. Thank you.

Answer 1

The LHS involves a geometric series: $$\sum_{k=0}^{n-2} 2^k = \frac{1-2^{n-1}}{1-2} = 2^{n-1}-1$$

Now multiply: $$(2^{n-1}+1)(2^{n-1}-1) = 4^{n-1}-1$$