Consider a function f:ℕ→ℝ, satisfying f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x≥2 with f(1)=1. Then 1/f(2022)+1/f(2028) is equal to

Solution

Maths sir · 4 steps

Write the relation for x and for x − 1, subtract, and the whole sum on the left collapses to one term.

1. Find f(2) first. Put x = 2: f(1) + 2f(2) = 6f(2). With f(1) = 1 this gives 4f(2) = 1, so f(2) = 1/4.
2. Subtract consecutive cases. For x ≥ 3 the relation holds at both x and x - 1. Subtracting them leaves only xf(x) on the left side.

   xf(x) = x(x+1)f(x) - (x-1)x f(x-1) ⇒ xf(x) = (x-1)f(x-1)

3. A constant product. So xf(x) takes the same value for every x ≥ 2, and that value is 2f(2) = 1/2. Hence f(x) = 1/2x, which means 1/f(x) = 2x for x ≥ 2.
4. So. 1/f(2022) + 1/f(2028) = 2(2022 + 2028) = 2 × 4050 = 8100, which is option (c).

The idea to keep. When a running sum equals an expression in f(x), write the relation for x and x-1 and subtract. Then fix the constant from the smallest x where the relation actually holds.

Answer from the NTA final answer key. Solution by PrepWiser.