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B

- Hướng dẫn giải

Đáp án B

$\begin{array}{l}\frac{1}{C_2^2}+\frac{1}{C_3^2}+\frac{1}{C_4^2}+...+\frac{1}{C_n^2}=\frac{9}{5}\\ \iff 1+\frac{1}{3}+\frac{1}{6}+\mathrm{...}+\frac{2}{n(n-1)}=\frac{9}{5}\\ \iff \frac{2}{2.3}+\frac{2}{3.4}+\mathrm{...}+\frac{2}{n(n-1)}=\frac{4}{5}\\ \iff \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{n-1}-\frac{1}{n}=\frac{2}{5}\\ \iff \frac{1}{2}-\frac{1}{n}=\frac{2}{5}\\ \iff \frac{1}{n}=\frac{1}{10}\\ \iff n=10\end{array}$