JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 21)
Suppose $$f$$ is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x,y \in N$$ and $$f(1) = {1 \over 5}$$. If $$\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}} $$, then $$m$$ is equal to __________.
Answer
10
Explanation
$\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}$
$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5}$
$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
$\therefore m=10$
$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5}$
$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
$\therefore m=10$
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