Let α=Σk=0ⁿ(( ⁿ Cₖ)²/(k+1)) and β=Σk=0ⁿ⁻¹((ⁿ Cₖ ⁿ Cₖ₊₁)/(k+2)) If 5 α=6 β, then n equals _______.

Solution

Maths sir · 4 steps

Absorb each denominator into a binomial coefficient, and Vandermonde collapses both sums.

1. Absorb the denominator. Use 1/(k+1) ⁿCₖ = 1/(n+1) ⁿ⁺¹Cₖ₊₁. In the same way, 1/(k+2) ⁿCₖ₊₁ = 1/(n+1) ⁿ⁺¹Cₖ₊₂. Also write the other factor as ⁿCₖ = ⁿCₙ₋ₖ.

   α = 1/(n+1)Σₖ₌₀ⁿ ⁿ⁺¹Cₖ₊₁ ⁿCₙ₋ₖ, β = 1/(n+1)Σₖ₌₀ⁿ⁻¹ ⁿ⁺¹Cₖ₊₂ ⁿCₙ₋ₖ

2. Apply Vandermonde. In each product the top numbers add to 2n+1. The bottom numbers add to a fixed value: (k+1)+(n-k) = n+1 in α, and (k+2)+(n-k) = n+2 in β. So α = 1/(n+1) ²ⁿ⁺¹Cₙ₊₁ and β = 1/(n+1) ²ⁿ⁺¹Cₙ₊₂.
3. Take the ratio. Use ^NCᵣ₊₁/^NCᵣ = (N-r)/(r+1) with N = 2n+1 and r = n+1. This gives β/α = (2n+1-(n+1))/(n+2) = n/(n+2).
4. So. 5α = 6β means β/α = 5/6. Then n/(n+2) = 5/6, so 6n = 5n + 10 and n = 10.

The idea to keep. A 1/(k+1) next to ⁿCₖ becomes 1/(n+1)ⁿ⁺¹Cₖ₊₁. After that, if the lower indices of a product add to a constant, Vandermonde collapses the sum.

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