Mathematics - Functions Question with Solution | TestHub

Given $\mathrm{f}(1)+2\mathrm{f}(2)+3\mathrm{f}(3)+\dots +\mathrm{nf}(\mathrm{n})=\mathrm{n}(\mathrm{n}+1)\mathrm{f}(\mathrm{n})$ ........(1)

Replacing $n$ by ( $n+1$ ), we get $\mathrm{f}(1)+2\mathrm{f}(2)+3\mathrm{f}(3)+\dots +\mathrm{nf}(\mathrm{n})+(\mathrm{n}+1)\mathrm{f}(\mathrm{n}+1)=(\mathrm{n}+1)(\mathrm{n}+2)\mathrm{f}(\mathrm{n}+1)$.(2) Subtracting (1) from (2), we get

$$\begin{array}{rlrl}& (n+1)f(n+1)=(n+1)(n+2)f(n+1)-n(n+1)f(n)& & \end{array}$$

$$\Rightarrow f(n+1)=(n+2)f(n+1)-nf(n)$$

$$\Rightarrow nf(n)=(n+2)f(n+1)-f(n+1)=(n+2-1)f(n+1)=(n+1)f(n+1)$$

Thus, we have $2f(2)=3f(3)=\dots =nf(n)$.

Hence from (1), $\mathrm{f}(1)+(\mathrm{n}-1)\mathrm{nf}(\mathrm{n})=\mathrm{n}(\mathrm{n}+1)\mathrm{f}(\mathrm{n})$

$$\begin{array}{rlrl}& \Rightarrow f(n)\{n(n+1)-n(n-1)\}=f(1)=1& & \end{array}$$

$$\Rightarrow \left(n^2+n-n^2+n\right)f(n)=1\Rightarrow 2nf(n)=1\Rightarrow f(n)=\frac{1}{2n}\therefore f(200)=\frac{1}{2\times 200}=\frac{1}{400}$$