Question

It is given that $f\left(1\right)+2f\left(2\right)+...+nf\left(n\right)=n(n+1)f\left(n\right)$.The value of$f\left(1\right)=1$and $f\left(999\right)$is$\frac{1}{k}$ , where $k$ equals.

Answer 1

Find the value of $k:$

Given that $f\left(999\right)=\frac{1}{k}$.

The summation of series will be utilized. We are aware that the formula for the function f is provided as

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)\dots +nf\left(n\right)=n(n+1)f\left(n\right)$…….(1)

If we use $n=n+1$ in the equation above, we get,

$f\left(1\right)+2f\left(2\right)+3f\left(3\right)\dots +nf\left(n\right)+(n+1)f(n+1)=(n+1)(n+2)f(n+1)\dots \dots \left(ii\right)$

Now, Subtract the both equations

$$\begin{gathered} =(n+1)f(n+1)=(n+1)(n+2)f(n+1)-n(n+1)f\left(n\right) \\ =(n+1)(n+2)f(n+1)-(n+1)f(n+1)=n(n+1)f\left(n\right) \\ =f(n+1)\left\{\right(n+1\left)\right(n+2)-(n+1\left)\right\}=n(n+1)f\left(n\right) \\ =(n+1)f(n+1)\{n+1\}=n(n+1)f\left(n\right) \\ =(n+1{)}^{2}f(n+1)=n(n+1\left)f\right(n) \end{gathered}$$

Remove $\left(n+1\right)$ from both sides of the equation, we get,

$\Rightarrow (n+1)f(n+1)=nf\left(n\right)$

Now for any n, we have,$(n+1)f(n+1)=nf\left(n\right)$.

This type of situation is only possible if the value of $nf\left(n\right)$ is a constant.

Let the constant be $C$, then $nf\left(n\right)=C$.
$\Rightarrow kf\left(k\right)=C$ where $C$ is a constant.
When $k=1$,
$\Rightarrow 1f\left(1\right)=C$
Now,$f\left(999\right)=\frac{1}{k}$, and we have,

$kf\left(k\right)=C$
$$\begin{gathered} \Rightarrow f\left(k\right)=\frac{C}{k} \\ \Rightarrow f\left(k\right)=\frac{1}{k}\text{when C}=1 \\ \Rightarrow k=999 \end{gathered}$$
Therefore , the required value of $k$ is $999$.