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QJEEM23Maf5bd: JEE Main 2023 Mathematics - Class 11 JEEM23Maf5bd Solution: Suppose is a function satisfying for all and . If , then is…

Suppose ff is a function satisfying f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) for all x,y∈Nx,y \in N and f(1)=15f(1) = {1 \over 5} . If ∑n=1mf(n)n(n+1)(n+2)=112\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}} , then mm is equal to __________.

Step-by-Step Verified Solution

∵f(1)=15 ∴f(2)=f(1)+f(1)=25\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}

f(2)=25f(3)=f(2)+f(1)=35f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}

f(3)=35f(3)=\frac{3}{5}

∴∑n=1mf(n)n(n+1)(n+2)\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}

=15∑n=1m(1n+1−1n+2)=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)

=15(12−13+13−14+….+1m+1−1m+2)=\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)

=15(12−1m+2)=m10(m+2)=112=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}

∴m=10\therefore m=10

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