1. Tardigrade
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  4. Let f: N arrow R be such that f(1) = 1 and f(1) + 2f(2) + 3f(3) + ... ..... + nf(n) = n(n + 1)f(n), for all n ∈ 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

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Q. Let $f : N \rightarrow R$ 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

WBJEEWBJEE 2015Relations and Functions - Part 2

Solution:

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