Uzasadnij tożsamość krok po kroku

[unikat900]

\(\displaystyle{ \sum_{k=1}^{n}k {n\choose k}=n2^{n-1}}\)

[mol_ksiazkowy]

\(\displaystyle{ f(x)=(1+x)^n}\)
\(\displaystyle{ s=f^{\prime}(1)}\)

[unikat900]

Do czego sie to odnosi

s = f'(1)

[mol_ksiazkowy]

\(\displaystyle{ f^\prime(x)=n(x+1)^{n-1}}\)

[Sylwek]

A po co tu pochodne angażować , ja bym to zrobił tak:

\(\displaystyle{ {n \choose k}={n \choose n-k} \\ S=\sum_{k=1}^{n}k {n\choose k}=\sum_{k=0}^{n}k {n\choose k} \\ \begin{cases}S=0 {n \choose 0} + 1 {n \choose 1} + \ldots + n {n \choose n} \\ S=n {n \choose 0} + (n-1) {n \choose 1} + \ldots + 0 {n \choose n} \end{cases} \\ 2S=(0+n) {n \choose 0} + (1+(n-1)) {n \choose 1} + \ldots + (n+0) {n \choose n}=n \sum_{k=0}^{n} {n\choose k}}\)

Ale z rozwinięcia dwumianu Newtona:
\(\displaystyle{ \sum_{k=0}^{n} {n\choose k}=(1+1)^n=2^n}\)

Zatem:
\(\displaystyle{ 2S=n 2^n \\ S=n 2^{n-1}}\)