Question

Let $f:N\to R$ be such that $f\left(1\right)=1$ and $f\left(1\right)+2f\left(2\right)+3f\left(3\right)+...+nf\left(n\right)=n(n+1)f\left(n\right)$, for all $n\in N,n\ge 2$, where N is the set of natural numbers and R is the set of real numbers. Then , the value of $f\left(500\right)$ is

• A. $1000$
• B. $500$
• C. $\frac{1}{500}$
• D. $\frac{1}{1000}$

Answer 1

Answer: (D)

$\frac{1}{1000}$

We have, $f:N\to R$ such that $f\left(1\right)=1$

And $f\left(1\right)+2f\left(2\right)+3f\left(3\right)++nf\left(n\right)=n(n+1)f\left(n\right),\forall n\ge 2$

Clearly, $f\left(1\right)+2f\left(2\right)=2(2+1)f\left(2\right)$

$\Rightarrow f\left(1\right)=6f\left(2\right)-2f\left(2\right)$

$\Rightarrow f\left(1\right)=4f\left(2\right)$

$\Rightarrow f\left(2\right)=\frac{f\left(1\right)}{4}=\frac{1}{4}$

Similarly, $f\left(1\right)+2f\left(2\right)+3f\left(3\right)=3(3+1)f\left(3\right)$

$\Rightarrow 1+\frac{1}{2}+3f\left(3\right)=12f\left(3\right)$

$\Rightarrow 9f\left(3\right)=\frac{3}{2}\Rightarrow f\left(3\right)=\frac{1}{6}$

and $f\left(1\right)+2f\left(2\right)+3f\left(3\right)+4f\left(4\right)=4\left(5\right)f\left(4\right)$

$\Rightarrow 1+\frac{1}{2}+\frac{1}{2}=16f\left(4\right)$

$\Rightarrow f\left(4\right)=\frac{2}{16}=\frac{1}{8}$

In general, $f\left(n\right)=\frac{1}{2n}$

$\therefore f\left(500\right)=\frac{1}{1000}$