Prove Prove $n2^{n - 1}$ = $\sum_{i\geq 0}^{} {n \choose i}i$ algebraically and using induction.

Question

Suppose $n \in$ natural numbers.

Prove $n2^{n - 1}$ = $\sum_{i\geq 0}^{} {n \choose i}i$

I have proven it combinatorially. Just having troubles algebraically and using induction.

Answer 1

Hint: Note that $\binom{n+1}{i+1}(i+1)=\binom{n}{i}(n+1)$.

Answer 2

For $n = 0$, both sides are $0$; assume the result holds for some $n$. Then we have \begin{align*} \sum_{i \geq 0 } \binom{n+1}{i}i &= \sum_{i\geq 0} \binom{n}{i}i + \sum_{i \geq 0} \binom{n}{i-1}i \\ &= n 2^{n-1} + 2^n + \sum_{i \geq 0} \binom{n}{i-1}(i-1) \\ &= n 2^{n-1} + 2^n + n 2^{n-1} \\ &= (n+1)2^n. \end{align*}