Question

urgent !! este dintr-o varianta de bac 2010

Answer 1

Salut,

$k\cdot C_n^k=k\cdot\dfrac{n!}{k!\cdot (n-k)!}=k\cdot\dfrac{n!}{k\cdot (k-1)!\cdot (n-k)!}=\\\\=\dfrac{n!}{(k-1)!\cdot (n-k)!}=\dfrac{n\cdot (n-1)!}{(k-1)!\cdot (n-k)!}=n\cdot\dfrac{(n-1)!}{(k-1)!\cdot (n-k)!}=\\\\=n\cdot\dfrac{(n-1)!}{(k-1)!\cdot [n-1-(k-1)]!}=n\cdot C_{n-1}^{k-1}.\ Suma\ din\ enun\c{t}\ devine:\\\\\sum\limits_{k=1}^n n\cdot C_{n-1}^{k-1}=n\cdot \sum\limits_{k=1}^n C_{n-1}^{k-1}=n\cdot(C_{n-1}^0+C_{n-1}^1+C_{n-1}^2+\ldots+C_{n-1}^{n-1})=\\\\=n\cdot(1+1)^{n-1}=n\cdot 2^{n-1}.\ Baft\breve{a}\ mult\breve{a}\ la\ examene\ !$

Green eyes.

Answer 2

$k C_n^k = k\cdot \dfrac{n!}{k!\cdot (n-k)!} = k\cdot \dfrac{n!}{(k-1)!\cdot k\cdot (n-k)!} = \\ \\ = \dfrac{n!}{(k-1)!\cdot (n-k)!} = \dfrac{(n-1)!\cdot n}{(k-1)!\cdot (n-k)!} = \\ \\ = n\cdot \dfrac{(n-1)!}{(k-1)!\cdot \Big((n-1)-(k-1)\Big)} = n\cdot C_{n-1}^{k-1} \\ \\ \\ \sum\limits_{k=1}^n k C_n^k = \sum\limits_{k=1}^n n\cdot C_{n-1}^{k-1} = n\cdot \sum\limits_{k=1}^nC_{n-1}^{k-1} =$

$= n\cdot \Big(\underset{\text{Suma coeficientilor binomiali}}{\underbrace{C_{n-1}^0+C_{n-1}^1+C_{n-1}^2+...+C_{n-1}^{n-1}}}\Big) = n\cdot 2^{n-1}$