JEE Mathematics 2023 Question with Solution

Given:

$$\sum_{k=1}^{x} k f(k) = x(x + 1)f(x), \quad x \geq 2$$

and $f(1)=1$.

Find: $f(2022)$.

From the relation for $x=n$,

$$\sum_{k=1}^{n} kf(k)=n(n+1)f(n)$$

and for $x=n-1$,

$$\sum_{k=1}^{n-1} kf(k)=(n-1)nf(n-1).$$

Subtracting the second equation from the first,

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

Dividing by $n$,

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

So,

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

and hence

$$f(n)=\frac{n-1}{n}f(n-1).$$

Now using $f(1)=1$,

$$f(2)=\frac{1}{2}f(1)=\frac{1}{2}, \quad f(3)=\frac{2}{3}f(2)=\frac{1}{3}.$$

This suggests

$$f(n)=\frac{1}{n}.$$

Indeed, repeated application gives

$$f(n)=\frac{n-1}{n}\cdot \frac{n-2}{n-1}\cdots \frac{1}{2}f(1)=\frac{1}{n}.$$

Therefore,

$$f(2022)=\frac{1}{2022}.$$

the solution states option A, but the extracted working leads to $f(2022)=\frac{1}{2022}$, which does not match any listed option. Among the given options, the source marks A as correct.

Write the defining equation at two consecutive values:

$$\sum_{k=1}^{n} kf(k)=n(n+1)f(n)$$ $$\sum_{k=1}^{n-1} kf(k)=(n-1)nf(n-1).$$

Subtracting removes the summation terms up to $n-1$ and leaves only $nf(n)$ on the left:

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

Now divide by $n$:

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

Rearranging,

$$nf(n)=(n-1)f(n-1).$$

Thus,

$$f(n)=\frac{n-1}{n}f(n-1).$$

Iterating this recurrence from $f(1)=1$,

$$f(n)=\frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{1}{2}\cdot 1=\frac{1}{n}.$$

So the value is $\frac{1}{2022}$.