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1.Relation and Function
hard
Suppose $f$ is a function satisfying $f ( x + y )= f ( x )+ f ( y )$ for all $x , y \in N$ and $f (1)=\frac{1}{5}$. If $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to $...............$.
A
$11$
B
$12$
C
$10$
D
$13$
(JEE MAIN-2023)
Solution
$\because f(1)=\frac{1}{5} \therefore f(2)=f(1)+f(1)=\frac{2}{5}$
$f(2)=\frac{2}{5} \quad f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5}$
$\therefore \sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}$
$=\frac{1}{5} \sum \limits_{n=1}^m\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
$\therefore m=10$
Standard 12
Mathematics
