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माना $f$ एक फलन है जो सभी $x, y \in \mathbb{N}$ के लिए $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})$ को संतुष्ट करता है एवं $\mathrm{f}(1)=\frac{1}{5}$ है यदि $\sum_{\mathrm{n}=1}^{\mathrm{m}} \frac{\mathrm{f}(\mathrm{n})}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{1}{12}$ हैं, तब $\mathrm{m}$ बराबर है_________.
$11$
$12$
$10$
$13$
Solution
$\because f(1)=\frac{1}{5} \therefore f(2)=f(1)+f(1)=\frac{2}{5}$
$f(2)=\frac{2}{5} \quad f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5}$
$\therefore \sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}$
$=\frac{1}{5} \sum \limits_{n=1}^m\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
$\therefore m=10$
