Simplify $0 \binom{n}{0} + 1 \binom{n}{1} + 2 \binom{n}{2} + \dots + n \binom{n}{n},$ where $n \ge 1.$
Enter your answer in the form $f(n) 2^{g(n)},$ where $f(n)$ and $g(n)$ are polynomials in $n$ with integer coefficients.
I know that $$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n .$$ But I'm not sure how to find the answer if there are numbers multiplied to each of the binoms.
Note that $$ \binom{n}{k}\binom{k}{1}=n\binom{n-1}{k-1}\quad (1\leq k\le n) $$ whence $$ \sum_{k=1}^nk\binom{n}{k}=\sum_{k=1}^nn\binom{n-1}{k-1}=n\sum_{k=1}^n\binom{n-1}{k-1}=n2^{n-1} $$ by the sum you referenced.
Since $$(1+x)^n= \sum_{r=0}^n {n\choose r}x^r$$ Differentiating both sides w.r.t. $x$ gives, $$n(1+x)^{n-1}= \sum_{r=0}^n r{n\choose r}x^{r-1}$$
Put $x=1$ and we get the required sum for $$0 \binom{n}{0} + 1 \binom{n}{1} + 2 \binom{n}{2} + \dots + n \binom{n}{n}= \ n2^{n-1}$$
