If Σr=1³⁰ r²( ³⁰ Cᵣ)²/(³⁰ Cᵣ₋₁)=α × 2²⁹, then α is equal to _________.

Solution

Maths sir · 4 steps

First simplify the ratio of neighbouring coefficients; the sum then turns into two standard sums.

1. Ratio of neighbours. For consecutive coefficients, ⁿCᵣ/ⁿCᵣ₋₁ = (n-r+1)/r. With n = 30 this is (31-r)/r.
2. Simplify each term. Write the term as r² · ³⁰Cᵣ · ³⁰Cᵣ/³⁰Cᵣ₋₁ = r² · ³⁰Cᵣ · (31-r)/r = r(31-r) ³⁰Cᵣ. So the sum is 31Σ r ³⁰Cᵣ - Σ r² ³⁰Cᵣ.

   Σᵣ r ⁿCᵣ = n 2ⁿ⁻¹, Σᵣ r² ⁿCᵣ = n(n+1) 2ⁿ⁻²

3. Put in numbers. With n = 30, Σ r ³⁰Cᵣ = 30 × 2²⁹ and Σ r² ³⁰Cᵣ = 30 × 31 × 2²⁸ = 15 × 31 × 2²⁹. The term for r = 0 is zero, so starting the sums at r = 1 changes nothing.
4. So. The sum is 31 × 30 × 2²⁹ - 15 × 31 × 2²⁹ = 31 × 15 × 2²⁹ = 465 × 2²⁹, so α = 465.

The common slip: 495. Taking Σ r² ³⁰Cᵣ as 30 × 29 × 2²⁸ forgets the extra Σ r ³⁰Cᵣ from writing r² = r(r-1) + r. It gives α = 930 - 435 = 495, not 465.

The idea to keep. When a sum has ⁿCᵣ divided by ⁿCᵣ₋₁, replace the ratio with (n-r+1)/r first. Then split r² as r(r-1) + r to use the standard sums.

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