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

The correct option is D $\frac{1}{1000}$

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+...+nf\left(n\right)=n(n+1)f\left(n\right)$
$\Rightarrow f\left(1\right)+2f\left(2\right)+3f\left(3\right)+...+(n-1)f(n-1)=(n-1)nf(n-1)$

Subtract the above series,
$nf\left(n\right)=(n-1)f(n-1)$
$\Rightarrow \frac{f\left(n\right)}{f(n-1)}=\frac{n-1}{n}$

$\Rightarrow \frac{f\left(n\right)}{f(n-1)}\times \frac{f(n-1)}{f(n-2)}\times \cdots =\frac{n-1}{n}\times \frac{n-2}{n-1}\times \cdots$
$\Rightarrow f\left(n\right)=\frac{1}{2n}$
Thus, $f\left(500\right)=\frac{1}{1000}$