Sets, Relations & Functions
Recursive Functional Equations
nta_pyq_2023_jan
Grade 11
Question:
Consider a function $f: \mathbb{N} \to \mathbb{R}$, satisfying $f(1) + 2f(2) + 3f(3) + \cdots + xf(x) = x(x+1)f(x)$; $x \ge 2$ with $f(1)=1$. Then $\dfrac{1}{f(2022)} + \dfrac{1}{f(2028)}$ is equal to
Step-by-Step Solution
Key Concept: Subtract the recurrence at $x$ and $x-1$ to get $xf(x) = \frac{1}{2}$ for $x \ge 2$, so $f(x)=\frac{1}{2x}$.
$f(x)=\frac{1}{2x}$ for $x\ge2$. $\frac{1}{f(2022)}=4044$, $\frac{1}{f(2028)}=4056$. Sum $=8100$.
Correct Answer: 4