Question

Să se calculeze suma: subpunctul b) - ex. A9
Clasa a 10-a , combinări

Answer 1

$b)~~S_n = C_{n}^1-2C_{n}^2+3C_{n}^3-...+(-1)^{n+1}nC_n^n \\ \\ kC_{n}^k = k\dfrac{n!}{k!(n-k)!} = \dfrac{n!}{(k-1)!(n-k)!} = \\ \\ = \dfrac{(n-1)!\,n}{(k-1)!\Big((n-1)-(k-1)\Big)!} = nC_{n-1}^{k-1} \\ \\ S_n = nC_{n-1}^0-nC_{n-1}^1+nC_{n-1}^{2}-...+(-1)^{n+1}nC_{n-1}^{n-1} \\ \\ S_n = n\Big(C_{n-1}^0-C_{n-1}^1+C_{n-1}^2-...+(-1)^{n-1+2}C_{n-1}^{n-1}\Big) \\ \\ S_n = n\Big(C_{n-1}^0-C_{n-1}^1+C_{n-1}^2-...+(-1)^{n-1}C_{n-1}^{n-1}\Big)$

$(1+x)^n = C_{n}^0+C_{n}^1x+C_{n}^2x^2+...+C_{n}^nx^n\\ \\ (1+x)^{n-1} = C_{n-1}^0+C_{n-1}^1x+C_{n-2}^2x^2+...+C_{n-1}^{n-1}x^{n-1} \\ \\ x = -1 \Rightarrow (1-1)^{n-1} = C_{n-1}^0-C_{n-1}^1+C_{n-1}^2-...+(-1)^{n-1}C_{n-1}^{n-1} \\ \\ \Rightarrow C_{n-1}^0-C_{n-1}^1+C_{n-1}^2-...+(-1)^{n-1}C_{n-1}^{n-1} = 0\\ \\ S_n = n\cdot 0 = \boxed{0}$