Suppose f is a function satisfying f ( x + y ) = f ( x ) + f ( y ) for all x , y and f ( 1 ) = 1 5 . If n = 1 m f ( n ) n ( n + 1 ) ( n + 2 ) = 1 12 , then m is equal to ___________ . [2023]

Question

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 $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{12},$ then $m$ is equal to ___________ . [2023]

Smart Achievers · Accepted Answer

(10)

Given $f (x + y) = f (x) + f (y)$

So, $f (2) = f (1) + f (1) = \frac{2}{5}; f (3) = f (2) + f (1) = \frac{3}{5}$

$∴ \sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{5} \sum_{n = 1}^{m} (\frac{1}{n + 1} - \frac{1}{n + 2})$

$= \frac{1}{5} (\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{m + 1} - \frac{1}{m + 2})$

$= \frac{1}{5} (\frac{1}{2} - \frac{1}{m + 2}) = \frac{m}{10 (m + 2)} = \frac{1}{12}$

$∴ m = 10$

Solution

Q.
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 $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{12},$ then $m$ is equal to ___________ . [2023]

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Solution

Q. Suppose f is a function satisfying f(x+y)=f(x)+f(y) for all x,y∈ℕ and f(1)=15. If ∑n=1mf(n)n(n+1)(n+2)=112, then m is equal to ___________ . [2023]

Ans. (10) Given f(x+y)=f(x)+f(y) and f(1)=15 So, f(2)=f(1)+f(1)=25; f(3)=f(2)+f(1)=35 ∴∑n=1mf(n)n(n+1)(n+2)=15∑n=1m(1n+1-1n+2) =15(12-13+13-14+…+1m+1-1m+2) =15(12-1m+2)=m10(m+2)=112 ∴m=10

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Q.
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 $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{12},$ then $m$ is equal to ___________ . [2023]