Chứng minh rằng: (n + 1)/(n + 2).[1/kC(n +1) + 1/(k + 1)C(n + 1) = 1/kCn 22/08/2021 Bởi Charlie Chứng minh rằng: (n + 1)/(n + 2).[1/kC(n +1) + 1/(k + 1)C(n + 1) = 1/kCn
Xét vế trái ta có $\dfrac{n+1}{n+2} \left( \dfrac{k! (n-k+1)!}{(n+1)!} + \dfrac{(k+1)! (n-k)!}{(n+1)!} \right) = \dfrac{1}{n+2} . \dfrac{k! (n-k+1)! + (k+1)! (n-k)!}{n!}$ $= \dfrac{1}{n+2} . \dfrac{k! . (n-k)! [(n-k+1 + k + 1]}{n!}$ $= \dfrac{k! (n-k)!}{n!} . \dfrac{1}{n+2} . (n+2)$ $= \dfrac{1}{C_n^k} = VP$ Bình luận
Xét vế trái ta có
$\dfrac{n+1}{n+2} \left( \dfrac{k! (n-k+1)!}{(n+1)!} + \dfrac{(k+1)! (n-k)!}{(n+1)!} \right) = \dfrac{1}{n+2} . \dfrac{k! (n-k+1)! + (k+1)! (n-k)!}{n!}$
$= \dfrac{1}{n+2} . \dfrac{k! . (n-k)! [(n-k+1 + k + 1]}{n!}$
$= \dfrac{k! (n-k)!}{n!} . \dfrac{1}{n+2} . (n+2)$
$= \dfrac{1}{C_n^k} = VP$