find-k-0-n-k-C-n-k-

Question Number 33072 by prof Abdo imad last updated on 10/Apr/18

f i n d \sum_{k = 0}^{n} k C_{n}^{k} .

Commented by abdo imad last updated on 10/Apr/18

l e t c o n s i d e r t h e p o l y n o m e p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k}

p (x) = {(x + 1)}^{n} a n d p^{^{'}} (x) = n {(x + 1)}^{n - 1} a n d f r o m

a n o t h e r s i d e p^{^{'}} (x) = \sum_{k = 1}^{n} k C_{n}^{k} x^{k - 1} \Rightarrow

p^{^{'}} (1) = \sum_{k = 1}^{n} k C_{n}^{k} = n 2^{n - 1} .

Answered by sma3l2996 last updated on 10/Apr/18

k^{n} C_{k} = \frac{n!}{(k - 1) (n - k)!} = \frac{n (n - 1)!}{(k - 1)! (n - 1 - (k - 1))!} = n^{n - 1} C_{k - 1}

S o \underset{k = 0}{\sum^{n}} k^{n} C_{k} = n \underset{k = 1}{\sum^{n}}^{n - 1} C_{k - 1}

W e h a v e \underset{k = 0}{\sum^{n}}^{n} C_{k} = 2^{n}

l e t l = k - 1

S o \underset{k = 1}{\sum^{n}}^{n - 1} C_{k - 1} = \underset{l = 0}{\sum^{n - 1}}^{n - 1} C_{l} = 2^{n - 1}

\underset{k = 0}{\sum^{n}} k^{n} C_{k} = 2^{n - 1} n