The question asked to find the solution to $ \sum_{k=1}^{n}(2k+1)\binom{n}{k}$. Here's what I did: By the binomial theorem, we know that $ \sum_{k=1}^{n}\binom{n}{k}x^k = (1+x)^n$. Set $x=x_1^2$. Then $ \sum_{k=1}^{n}\binom{n}{k}(x_1^2)^k = \sum_{k=1}^{n}\binom{n}{k}x_1^{2k} = (1+x_1^2)^n$. If you multiply by $x_1$ and differentiate with respect to $x_1$, you get $\sum_{k=1}^{n}\binom{n}{k}x_1^{2k+1} = x_1(1+x_1^2)^n \implies \sum_{k=1}^{n}(2k+1)\binom{n}{k}x_1^{2k} = (1+x_1^2)^n + 2nx_1^2(1+x_1^2)^{n-1}$
So, $\sum_{k=1}^{n}(2k+1)\binom{n}{k} = 2^n + n2^{n} = 2^n(n+1)$ when $x_1=1$.
Whenever I test this by typing it into a calculator, it's always off by one, so I think the actual solution is $2^n(n+1)-1$, but I am not sure where I messed up. I would appreciate feedback.
Here’s an alternative approach that avoids derivatives: \begin{align} \sum_{k=1}^n (2k+1)\binom{n}{k} &= 2\sum_{k=1}^n k\binom{n}{k}+\sum_{k=1}^n \binom{n}{k} \\ &= 2\sum_{k=1}^n n\binom{n-1}{k-1}+\sum_{k=1}^n \binom{n}{k} \\ &= 2n\sum_{k=0}^{n-1} \binom{n-1}{k}+\sum_{k=1}^n \binom{n}{k} \\ &= 2n\cdot 2^{n-1} + (2^n-1) \\ &= (n+1)2^n-1 \end{align}