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1.Relation and Function
hard
ધારોકે $f$ એ પ્રત્યેક $f(x+y)=f(x)+f(y)$ માટે $x, y \in N$ અને $f(1)=\frac{1}{5}$ નું સમાધાન કરતુ વિધેય છે. જો $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$ હોય, તો $m=..........$
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
