Question

Consider a function $f:N\to R$, satisfying $f(1)+2f(2)+3f(3)+\dots +xf(x)=x(x+1)f(x);x\ge 2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

• A. 8200
• B. 8000
• C. 8400
• D. 8100

Answer 1

Answer: (D)

Given for $x\ge 2$
$f(1)+2f(2)+\dots \dots +xf(x)=x(x+1)f(x)$
$\text{ replace }x\text{ by }x+1$
$\Rightarrow x(x+1)f(x)+(x+1)f(x+1)$
$=(x+1)(x+2)f(x+1)$
$\Rightarrow \frac{x}{f(x+1)}+\frac{1}{f(x)}=\frac{(x+2)}{f(x)}$
$\Rightarrow xf(x)=(x+1)f(x+1)=\frac{1}{2},x\ge 2$
$f(2)=\frac{1}{4},f(3)=\frac{1}{6}$
$\text{ Now }f(2022)=\frac{1}{4044}$
$f(2028)=\frac{1}{4056}$
$\text{ So, }\frac{1}{f(2022)}+\frac{1}{f(2028)}=4044+4056=8100$