Let for x ∈ ℝ, S₀(x)=x, Sₖ(x)=Cₖ x+k ∫₀ˣ Sₖ₋₁(t) d t, where
C₀=1, Cₖ=1-∫₀¹ Sₖ₋₁(x) d x, k=1,2,3, … Then S₂(3)+6 C₃ is equal to ____________.

Solution

Maths sir · 3 steps

Build the chain one rung at a time: each integral of Sₖ₋₁ gives you both Cₖ and Sₖ.

1. The recipe. At each stage, integrate Sₖ₋₁ from 0 to 1 to get Cₖ = 1 - ∫₀¹ Sₖ₋₁ dx. Then integrate Sₖ₋₁ from 0 to x, multiply by k, and add Cₖ x to get Sₖ(x).

   k	Cₖ	Sₖ(x)	∫₀¹ Sₖ(x) dx
   0	1	x	1/2
   1	1 - 1/2 = 1/2	x/2 + x²/2	1/4 + 1/6 = 5/12
   2	1 - 5/12 = 7/12	7x/12 + x²/2 + x³/3	7/24 + 1/6 + 1/12 = 13/24
   3	1 - 13/24 = 11/24	not needed	not needed

2. Where S2 comes from. S₂(x) = 7/12x + 2∫₀ˣ (t/2 + t²/2)dt = 7x/12 + 2(x²/4 + x³/6). The factor k = 2 turns this into 7x/12 + x²/2 + x³/3.

   S₂(3) = 7/4 + 9/2 + 9 = 61/4, 6C₃ = 6 × 11/24 = 11/4

3. So. S₂(3) + 6C₃ = 61/4 + 11/4 = 72/4 = 18.

The common slip: 12. Dropping the factor k = 2 in front of the integral for S₂ gives S₂(3) = 8.5 and C₃ = 7/12, so the total comes out as 12.

The idea to keep. For a recursively defined integral sequence, do not hunt for a general formula. Compute the first few terms carefully in order, and keep every multiplier in the definition.

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