Let α=Σr=0ⁿ(4 r²+2 r+1) ⁿ Cᵣ and β=(Σr=0ⁿ (ⁿ Cᵣ)/(r+1))+1/(n+1). If 140<(2 α)/β<281, then the value of n is _________.

Solution

Maths sir · 4 steps

Split 4r²+2r+1 into the standard sums, and 2α/β collapses to a perfect cube.

1. Standard sums. Use Σ ⁿCᵣ = 2ⁿ, Σ r ⁿCᵣ = n 2ⁿ⁻¹ and Σ r² ⁿCᵣ = n(n+1) 2ⁿ⁻². So α = 4n(n+1)2ⁿ⁻² + 2n 2ⁿ⁻¹ + 2ⁿ = 2ⁿ[n(n+1) + n + 1] = 2ⁿ(n+1)².
2. Integrate for beta. Since ⁿCᵣ/(r+1) = ⁿ⁺¹Cᵣ₊₁/(n+1), the sum is (2ⁿ⁺¹-1)/(n+1). Adding 1/(n+1) gives β = 2ⁿ⁺¹/(n+1).

   2α/β = (2 × 2ⁿ (n+1)² × (n+1))/2ⁿ⁺¹ = (n+1)³

3. Check the cubes. 5³ = 125 is too small and 7³ = 343 is too large. Only 6³ = 216 lies between 140 and 281, so n + 1 = 6.
4. So. n = 5.

The common slip: 6. Reporting n + 1 = 6 instead of n gives 6. The cube is (n+1)³, so subtract 1 at the end.

The idea to keep. For sums of a polynomial in r times ⁿCᵣ, break the polynomial into 1, r, r² and use the standard results. For ⁿCᵣ/(r+1), write it as ⁿ⁺¹Cᵣ₊₁/(n+1).

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