If n(2 n+1) ∫₀¹(1-xⁿ)²ⁿ d x=1177 ∫₀¹(1-xⁿ)²ⁿ⁺¹ d x, then n ∈ N is equal to ______________.

Solution

Maths sir · 4 steps

Integrate by parts once, and the two integrals get tied by a clean ratio that you match to 1177.

1. Name the integrals. Let Iₘ = ∫₀¹ (1-xⁿ)ᵐ dx. The question compares I₂ₙ with I₂ₙ₊₁, so look for a relation between Iₘ and Iₘ₋₁.
2. By parts. Take (1-xⁿ)ᵐ as the first function and 1 as the second. The boundary term [x(1-xⁿ)ᵐ]₀¹ is 0, which leaves Iₘ = mn∫₀¹ xⁿ(1-xⁿ)ᵐ⁻¹ dx. Now write xⁿ = 1-(1-xⁿ), so the integral becomes Iₘ₋₁ - Iₘ.

   Iₘ = mn(Iₘ₋₁ - Iₘ) ⇒ (1+mn) Iₘ = mn Iₘ₋₁

3. Choose the power. Put m = 2n+1. This gives n(2n+1) I₂ₙ = (n(2n+1)+1) I₂ₙ₊₁. Comparing with the given equation, 2n² + n + 1 = 1177.
4. So. 2n² + n - 1176 = 0 gives n = (-1 ± √(1 + 9408))/4 = (-1 ± 97)/4. The natural number root is n = 24.

The idea to keep. For ∫₀¹ (1-xⁿ)ᵐ dx, integrate by parts and then replace xⁿ by 1-(1-xⁿ). You get (1+mn)Iₘ = mn Iₘ₋₁, which settles any question that links consecutive powers.

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