Let n∈N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms ⁿC₀,3.ⁿC₁,5.ⁿC₂,7.ⁿC₃,..... is equal to 2¹⁰⁰ . 101, then 2[ (n - 1)/2 ] is equal to _______________.

Solution

Maths sir · 4 steps

The coefficients are 2r+1, so split the sum into 2Σ r ⁿCᵣ plus Σ ⁿCᵣ and find n.

1. General term. The coefficients 1, 3, 5, 7, … go up by 2, so the r-th term is (2r+1) ⁿCᵣ for r = 0 to n. That gives exactly n+1 terms.
2. Two standard sums. Use Σ r ⁿCᵣ = n 2ⁿ⁻¹ and Σ ⁿCᵣ = 2ⁿ. The first comes from r ⁿCᵣ = n ⁿ⁻¹Cᵣ₋₁.

   S = 2(n 2ⁿ⁻¹) + 2ⁿ = (n+1) 2ⁿ

3. Find n. Set (n+1) 2ⁿ = 2¹⁰⁰ × 101. Matching gives n = 100.
4. So. [(n-1)/2] = [99/2] = [49.5] = 49, so 2 × 49 = 98.

The common slip: 99. Ignoring the greatest integer function and taking 2 × 99/2 gives 99. Take the floor of 49.5 first, then double.

The idea to keep. When the coefficients in front of ⁿCᵣ form an AP a + rd, the sum is a 2ⁿ + d n 2ⁿ⁻¹. Break it into the two standard sums.

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