A question regarding modified binomial summation

Question

We know that $$ \sum_{k=0}^N\binom {N} {k}a^N b^{N-k} = (a+b)^N $$ Do we have a formula for the following formula? $$ \sum_{k=0}^N k\cdot\binom {N} {k}a^{k} b^{N-k} $$ It looks very similar to compute the mean of a binomial distribution, but $a+b\neq 1$. Thanks!

Answer 1

There is a typo in the first line: it is: $$ \sum_{k=0}^{N}\binom{N}{k}a^\color{red}{k} b^{N-k}=(a+b)^N\tag{1} $$ so, by differentiating both sides with respect to $a$, then multiplying them by $a$: $$ \sum_{k=0}^{N}\binom{N}{k}k a^k b^{N-k} = Na(a+b)^{N-1}.\tag{2}$$