Chứng minh

Question

Chứng minh :

$\frac{n+1}{n+2}(\frac{1}{C^{k}_{n+1}}+ \frac{1}{C^{k+1}_{n+1}})=\frac{1}{C^{k}_{n}}$

score 1 · Answer 1

$VT=\dfrac{n+1}{n+2} \bigg ( \dfrac{k! (n-k+1)! + (k+1)! (n-k)!}{(n+1)!} \bigg )$

$=\dfrac{n+1}{n+2} \bigg ( \dfrac{k! (n-k)! \bigg [ n-k+1 +k+1 \bigg ] }{(n+1)!} \bigg )$

$=\dfrac{n+1}{n+2} . \dfrac{k!(n-k)! .(n+2)}{(n+1)!}= (n+1) .\dfrac{k! (n-k)!}{n! . (n+1)}=\dfrac{k! (n-k)!}{n!}=\dfrac{1}{C_n^k}$

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