<s>[INST] Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?

$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
 2)2^{n-1}$$

I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x = 1$, but I didn't get to my desired result. I just kept finding that $\displaystyle\sum_{k = 0}^n k \binom{n}{k} = n2^{n-1}$.
 [/INST] Here's an alternative proof that does not depend on derivatives:
\begin{align}
\sum_k (k+1) \binom{n}{k} 
&= \sum_k k\binom{n}{k} + \sum_k \binom{n}{k} \\
&= \sum_k n\binom{n-1}{k-1} + \sum_k \binom{n}{k} \\
&= n 2^{n-1} + 2^n \\
&= (n+2) 2^{n-1}
\end{align}
</s>