Suppose f is a function satisfying fx+y=fx+fy for all x,y∈ N &; F1=15 if ∑n=1mfnnn+1n+2=112 then m is equal to

fx+y=fx+ fy⇒fx=kxf1=15⇒k=15∑n=1mfnnn+1n+2=15∑1n+1n+2=1512−1m+2=m10m+2m10m+2=112⇒m=10

Suppose f is a function satisfying fx+y=fx+fy for all x,y∈ N &; F1=15 if ∑n=1mfnnn+1n+2=112 then m is equal to

fx+y=fx+ fy⇒fx=kxf1=15⇒k=15∑n=1mfnnn+1n+2=15∑1n+1n+2=1512−1m+2=m10m+2m10m+2=112⇒m=10

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Suppose $f$ is a function satisfying $f (x + y) = f (x) + f (y)$ for all $x, y \in N &$ ; $F (1) = \frac{1}{5}$ if $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{12}$ then $m$ is equal to

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Detailed Solution

$f (x + y) = f (x) + f (y) \Rightarrow f (x) = k x$

$f (1) = \frac{1}{5} \Rightarrow k = \frac{1}{5}$

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

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

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

$\Rightarrow m = 10$

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Suppose f is a function satisfying fx+y=fx+fy for all x,y∈ N &; F1=15 if ∑n=1mfnnn+1n+2=112 then m is equal to

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Suppose $f$ is a function satisfying $f (x + y) = f (x) + f (y)$ for all $x, y \in N &$ ; $F (1) = \frac{1}{5}$ if $\sum_{n = 1}^{m} \frac{f (n)}{n (n + 1) (n + 2)} = \frac{1}{12}$ then $m$ is equal to