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 (integer)
10
Solution
$\because f(1)=\frac{1}{5} ~\therefore f(2)=f(1)+f(1)=\frac{2}{5}$
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$f(2)=\frac{2}{5} \quad\quad f(3)=f(2)+f(1)=\frac{3}{5}$
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$f(3)=\frac{3}{5}$
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$\therefore \sum\limits_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$
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$=\frac{1}{5} \sum\limits_{n=1}^{m}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
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$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
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$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12}$
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$\therefore m=10$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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