Question

Let $f:N\to R$ be a function satisfying the following conditions:
$f\left(1\right)=1$ and $f\left(1\right)+2f\left(2\right)+\dots +nf\left(n\right)=n(n+1)f\left(n\right)$ for $n\ge 2$.
If $f\left(999\right)=\frac{1}{K}$, then $K$ equals

• A. $999$
• B. $1000$
• C. $1998$
• D. $2000$

Answer 1

The correct option is C $1998$

$f\left(1\right)+2f\left(2\right)+\dots +nf\left(n\right)=n(n+1)f\left(n\right)$ for $n\ge 2$
$\dots \left[1\right]$
Replacing $n$ by $n+1$, we get
$f\left(1\right)+2f\left(2\right)+\dots +nf\left(n\right)+(n+1)f(n+1)=(n+1)(n+2)f(n+1)$
$\dots \left[2\right]$
From $\left[2\right]-\left[1\right]$, we get
$(n+1)f(n+1)=(n+1)(n+2)f(n+1)-nf\left(n\right)$
$\Rightarrow f(n+1)=(n+2)f(n+1)-nf\left(n\right)$
$\Rightarrow (n+1)f(n+1)=nf\left(n\right)$
Putting $n=2,3,4,\dots ,$ we get
$2f\left(2\right)=3f\left(3\right)=4f\left(4\right)=\dots =nf\left(n\right)$
From $\left[1\right]$,
$f\left(1\right)+(n-1)nf\left(n\right)=n(n+1)f\left(n\right)$
$\Rightarrow f\left(1\right)=2nf\left(n\right)$
$\Rightarrow f\left(n\right)=\frac{f\left(1\right)}{2n}=\frac{1}{2n}$
$f\left(999\right)=\frac{1}{1998}$
$\therefore K=1998$