Question

Let $f$ be a function from the set of positive integers to the set of real number such that
(i) $f\left(1\right)=1$
(ii) $\sum _{r=1}^{n}rf\left(r\right)=n(n+1)f\left(n\right),\forall n\ge 2$
then find the value of $2126f\left(1063\right)$

Answer 1

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+...+n\left(f\right(n\left)\right)=n(n+1)f\left(n\right)$
$\Rightarrow f\left(1\right)+2\left(2\right)+3\left(3\right)+..+(n+1)f(n+1)=(n+1)(n+2)f(n+1)$
$\Rightarrow n(n+1)f\left(n\right)=(n+1{)}^{2}f(n+1)\Rightarrow nf(n)=(n+1\left)f\right(n+1)$
ie, $2f\left(2\right)=3f\left(3\right)=....=n\left(f\right(n\left)\right)$
ie, $f\left(n\right)=\frac{1}{n}$
$2126f\left(1063\right)=2$