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 $nϵN,n\ge 2$, 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. $1/500$
• D. $1/1000$

Answer 1

The correct option is D $1/1000$

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+.....+nf\left(n\right)=n(n+1)f\left(n\right)$
$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+.....+(n-1)f(n-1)=(n-1)nf(n-1)$
Subtracting, $nf\left(n\right)=n(n+1)f\left(n\right)-n(n-1)f(n-1)$
$\Rightarrow nf\left(n\right)=(n-1)f(n-1)$
Clearly, $f\left(n\right)=\frac{1}{2n}$. So, $f\left(500\right)=\frac{1}{1000}$