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

$$(n+1)f(n+1)=(n+1)(n+2)f(n+1)-n(n+1)f(n)$$

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

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

Thus, we have $2\mathrm{f}(2)=3\mathrm{f}(3)=\dots =\mathrm{nf}(\mathrm{n})$. Hence from (1), $\mathrm{f}(1)+(\mathrm{n}-1)\mathrm{nf}(\mathrm{n})=\mathrm{n}(\mathrm{n}+1)\mathrm{f}(\mathrm{n})$.

$$\mathrm{f}(\mathrm{n})\{\mathrm{n}(\mathrm{n}+1)-\mathrm{n}(\mathrm{n}-1)\}=\mathrm{f}(1)=1$$

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

$$2\mathrm{nf}(\mathrm{n})=1$$

$$\mathrm{f}(\mathrm{n})=\frac{1}{2\mathrm{n}}$$

$$\therefore \mathrm{f}(200)=\frac{1}{2\times 200}=\frac{1}{400}$$