Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically.

Question

Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically.

I know $2^n$ = $\sum_{i\geq 0}^{} {n\choose i}$. But how do I manipulate the $(n+1)$ to make it look like the right side?

Answer 1

Using the binomial theorem, we have

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

Taking polynomial derivative with respect to $x$, we have

$$(n+1)(1+x)^n=\sum_{i=0}^{n+1}{n+1\choose i}\,i\,x^{i-1}.$$

Setting $x=1$, we obtain

$$(n+1)2^n=\sum_{i=0}^{n+1}{n+1\choose i}\,i.$$

Only using derivatives of polynomials. Still algebraic?

Answer 2

$$ \begin{align} \sum_{k=0}^{n+1}\binom{n+1}{k}k &=\sum_{k=1}^{n+1}\binom{n+1}{k}k\\ &=\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{n+1}{k}\cdot k\\ &=\sum_{k=1}^{n+1}\binom{n}{k-1}(n+1)\\[6pt] &=(n+1)\,2^n \end{align} $$

Answer 3

Take the derivative of \begin{align} (1+x)^{n+1} = \sum_{i=0}^n \binom{n+1}{i} x^i \end{align} with respect to $x$ and then plug in $x=1$.