Question

Let $f:N\to R$ be such that $f(1)=1$ and
$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+........+nf\left(n\right)=n\left(n+1\right)f\left(n\right),$ 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 $f(500)$ is

• A. $100$
• B. $500$
• C. $1/500$
• D. $1/1000$

Answer 1

Answer: (D)

We have, $f:N\to R$ such that $f(1)=1$ a
nd $f(1)+2f(2)+3f(3)+\dots +nf(n)$
$=n(n+1)f(n),\forall n\ge 2$
Clearly, $f(1)+2f(2)=2(2+1)f(2)$
$\Rightarrow f(1)=6f(2)-2f(2)$
$\Rightarrow f(1)=4f(2)$
$\Rightarrow f(2)=\frac{f(1)}{4}=\frac{1}{4}$
Similarly, $f(1)+2f(2)+3f(3)=3(3+1)f(3)$
$\Rightarrow 1+\frac{1}{2}+3f(3)=12f(3)$
$\Rightarrow 9f(3)=\frac{3}{2}$
$\Rightarrow f(3)=\frac{1}{6}$
and $f(1)+2f(2)+3f(3)+4f(4)=4(5)f_1$,
$\Rightarrow 1+\frac{1}{2}+\frac{1}{2}=16f(4)$
$\Rightarrow f(4)=\frac{2}{16}=\frac{1}{8}$
In general, $f(n)=\frac{1}{2n}$
$\therefore f(500)=\frac{1}{1000}$