If Cᵣ ≡ ²⁵Cᵣ and
C₀ + 5.C₁ + 9.C₂ + .... + (101).C₂₅ = 2²⁵.k, then k is equal to _____.

Solution

Maths sir · 3 steps

Spot that the multipliers are 4r+1, then split the sum into two standard binomial sums.

1. Find the pattern. The multipliers 1, 5, 9, …, 101 go up by 4, so the coefficient of Cᵣ is 4r + 1. Check the last term: 4 × 25 + 1 = 101.
2. Split the sum. So the sum is 4Σ r Cᵣ + Σ Cᵣ. Use Σᵣ₌₀ⁿ Cᵣ = 2ⁿ and Σᵣ₌₀ⁿ r Cᵣ = n · 2ⁿ⁻¹ with n = 25.

   Σᵣ₌₀²⁵(4r+1) Cᵣ = 4 × 25 × 2²⁴ + 2²⁵ = 50 × 2²⁵ + 2²⁵ = 51 × 2²⁵

3. So. Comparing with 2²⁵ k gives k = 51.

The common slip: 101. Taking Σ r Cᵣ as n · 2ⁿ instead of n · 2ⁿ⁻¹ gives 100 × 2²⁵ + 2²⁵, so k = 101 instead of 51.

The idea to keep. When the multipliers of Cᵣ form an AP, write them as ar + b. Then you only need Σ Cᵣ = 2ⁿ and Σ r Cᵣ = n · 2ⁿ⁻¹.

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