Animated Solution for Mathematics - Sequence and Series: 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 $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

Functional equations often appear as daunting, abstract puzzles, but they are essentially stories about how a function evolves. Today, we are going to peel back the layers of a problem that seems to demand complex calculus but actually rewards elegant algebraic intuition.

We are given a function $f:\mathbb{N}\to \mathbb{R}$ with an initial condition $f(1)=1$ and a summation constraint:

Our goal is to find the value of $\frac{1}{f(2022)}+\frac{1}{f(2028)}$. Take a deep breath; we are going to solve this step-by-step.

The Summation Trap

When you see a summation like ${\sum }_{k=1}^{x}kf(k)$, your first instinct should be to simplify it. Let us define $S_x={\sum }_{k=1}^{x}kf(k)$.

The problem states that $S_x=x(x+1)f(x)$. Recall the fundamental property of any series: the sum of the first $x$ terms minus the sum of the first $x-1$ terms leaves you with exactly the $x$-th term.

Mathematically, this is expressed as:

This is our bridge. By substituting our definition of $S_x$ into this difference, we create a path to isolate $f(x)$.

The Recurrence Dance

Let us perform the substitution. We have $S_x=x(x+1)f(x)$ and $S_{x-1}=(x-1)xf(x-1)$.

Plugging these into our difference equation gives us:

Since we are dealing with $x\ge 2$, we know $x eq 0$. This allows us to divide the entire equation by $x$ without any fear of losing information.

The equation simplifies to:

Now, let us group the $f(x)$ terms. Moving $f(x)$ to the left, we get $(x+1)f(x)-f(x)=(x-1)f(x-1)$, which simplifies to $xf(x)=(x-1)f(x-1)$.

We have arrived at our recurrence relation:

Unrolling the Pattern

Every recurrence needs a starting point. We know $f(1)=1$, but our recurrence is valid for $x\ge 2$.

Let us find $f(2)$ using the original summation equation: $f(1)+2f(2)=2(2+1)f(2)$. Substituting $f(1)=1$, we get $1+2f(2)=6f(2)$, which leads to $4f(2)=1$, or $f(2)=\frac{1}{4}$.

Now, let us see what happens as we move forward:

For $x=3$:

For $x=4$:

The pattern is undeniable: $f(x)=\frac{1}{2x}$ for $x\ge 2$.

We have cracked the code. The function is $f(x)=\frac{1}{2x}$ for $x\ge 2$.

The question asks for $\frac{1}{f(2022)}+\frac{1}{f(2028)}$. Since $f(x)=\frac{1}{2x}$, the reciprocal is simply $\frac{1}{f(x)}=2x$.

Therefore:

Adding these together:

We have navigated the complexity and emerged with a clean, elegant result. Remember, in JEE Advanced, the most complex-looking problems often yield to the simplest, most fundamental principles. The final answer is 8100.