Q105 / 160 · WBJEE 2015 · Mathematics
Let $f: N \rightarrow R$ be such that $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n),$ for all $n \in N, n \geq 2,$ where $N$ is the set of natural numbers and $R$ is the set of real numbers. Then, the value of $f(500)$ is
- A1000
- B500
- C$1 / 500$
- Correct answerD$1 / 1000$
Solution
We have, $f: N \rightarrow R$ such that $f(1)=1$
$$
\begin{array}{l}
\text { and } f(1)+2 f(2)+3 f(3)+\ldots+n f(n) \\
\quad=n(n+1) f(n), \forall n \geq 2 \\
\text { Clearly, } f(1)+2 f(2)=2(2+1) f(2) \\
\Rightarrow \quad f(1)=6 f(2)-2 f(2) \\
\Rightarrow \quad f(1)=4 f(2) \\
\Rightarrow \quad f(2)=\frac{f(1)}{4}=\frac{1}{4}
\end{array}
$$
Similarly.
$$
f(1)+2 f(2)+3 f(3)=3(3+1) f(3)
$$
$\Rightarrow \quad 1+\frac{1}{2}+3 f(3)=12 f(3)$
$\Rightarrow \quad 9 f(3)=\frac{3}{2} \Rightarrow f(3)=\frac{1}{6}$
and $f(1)+2 f(2)+3 f(3)+4 f(4)=4(5) f(4)$
$\Rightarrow \quad 1+\frac{1}{2}+\frac{1}{2}=16 f(4)$
$\Rightarrow \quad f(4)=\frac{2}{16}=\frac{1}{8}$
In general, $f(n)=\frac{1}{2 n}$
$\therefore \quad f(500)=\frac{1}{1000}$
Hence, the correct answer is option: 'D'.