Question

Let f : $N\to R$ be a function such that $f\left(1\right)+2f\left(2\right)+3f\left(3\right).......+nf\left(n\right)=n(n+1)f\left(n\right)$, for $n\ge 2$ and $f\left(1\right)=1$ then

• A. $f\left(n\right)=\frac{1}{n}$
• B. $f\left(5\right)=\frac{1}{10}$
• C. $f\left(5\right)=\frac{1}{5}$
• D. $f\left(n\right)=\frac{1}{3n}$

Answer 1

The correct option is B $f\left(5\right)=\frac{1}{10}$

We have $f\left(1\right)+2f\left(2\right)+3f\left(3\right).......+nf\left(n\right)=n(n+1)f\left(n\right)$, for $n\ge 2$.......(1) and $f\left(1\right)=1$.

Putting $n=2$ in (1) and using the value of $f\left(1\right)$we get,

$1+2f\left(2\right)=6f\left(2\right)$

or, $4f\left(2\right)=1$

or, $f\left(2\right)=\frac{1}{2}$.

Again $f\left(3\right)=\frac{1}{6}$.

By induction it can be proved $f\left(n\right)=\frac{1}{2n}$ for $n\ge 2$ and $f\left(1\right)=1$.

Again $f\left(5\right)=\frac{1}{10}$.

So option (B) is correct.