Ta có: [tex]\frac{C_{n}^{k}}{C_{n+2}^{k+1}}=\frac{\frac{n!}{(n-k)!k!}}{\frac{(n+2)!}{(k+1)!(n-k+1)!}}=\frac{n!(k+1)!(n-k+1)!}{(n+2)!(n-k)!k!}=\frac{(n-k+1)(k+1)}{(n+1)(n+2)}=\frac{n(k+1)+1-k^{2}}{(n+1)(n+2)}(*)[/tex]
Áp dụng $(*)$ vào $S$ ta được:
$S$ [tex]=\frac{n.1+1-0^{2}}{(n+1).(n+2)}+\frac{n.2+1-1^{2}}{(n+1)(n+2)}+\frac{n.3+1-2^{2}}{(n+1).(n+2)}+\cdots +\frac{n.(n+1)+1-n^{2}}{(n+1).(n+2)}\\ =\frac{n.\left [ 1+2+3+\cdots+(n+1) \right ]+1.(n+1)-\left (1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right )}{(n+1)(n+2)}\\=\frac{n.\frac{(n+2)(n+1)}{2}}{(n+1)(n+2)}+\frac{1}{n+2}-\frac{\frac{n(n+1)(2n+1)}{6}}{(n+1)(n+2)}\\ =\frac{n}{2}+\frac{1}{n+2}-\frac{n(2n+1)}{6(n+2)}=\frac{6-2n^{2}-n+3n(n+2)}{6(n+2)}\\ =\frac{n^{2}+5n+6}{6(n+2)}=\frac{(n+2)(n+3)}{n+2}=\frac{n+3}{6}=\frac{n}{6}+\frac{1}{2}\\ \Rightarrow \begin{cases} a=6 \\ b= 2 \end{cases} \Rightarrow a+b=8[/tex]
Nếu thắc mắc chỗ nào thì cậu hỏi lại nhé! Chúc cậu học tốt! :3