Question

Let $f:N\to R$ be a function satisfying $f\left(1\right)+2f\left(2\right)+3f\left(3\right)+4f\left(4\right)+\dots +nf\left(n\right)=n(n+1)f\left(n\right)$ for $n\ge 2$. If $f\left(1\right)=1$, then which of the following is/are true?

• A. $f\left(1003\right)=\frac{1}{2006}$
• B. $f\left(999\right)=\frac{1}{1998}$
• C. $f\left(1998\right)=\frac{1}{1998}$
• D. $f\left(2\right),f\left(3\right)$ and $f\left(4\right)$ are in H.P.

Answer 1

Answer: (A), (B), (D)

$f\left(2\right),f\left(3\right)$ and $f\left(4\right)$ are in H.P.

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)+4f\left(4\right)+\dots +nf\left(n\right)=n(n+1)f\left(n\right)\Rightarrow f\left(1\right)+2f\left(2\right)+3f\left(3\right)+4f\left(4\right)+\dots +(n-1)f(n-1)=n^{2}f\left(n\right)\Rightarrow f\left(n\right)=\frac{\sum _{r=1}^{n-1}rf\left(r\right)}{n^2}$
Putting $n=2$
$f\left(2\right)=\frac{f\left(1\right)}{4}=\frac{1}{4}$
Putting $n=3$
$f\left(3\right)=\frac{f\left(1\right)+2f\left(2\right)}{9}=\frac{1}{6}$
Putting $n=4$
$f\left(4\right)=\frac{f\left(1\right)+2f\left(2\right)+3f\left(3\right)}{16}=\frac{1}{8}\therefore f\left(n\right)=\frac{1}{2n}$
$\Rightarrow f\left(1003\right)=\frac{1}{2006},f\left(999\right)=\frac{1}{1998},f\left(1998\right)=\frac{1}{3996}$
Also, $f\left(2\right),f\left(3\right)$ and $f\left(4\right)$ are in H.P.