Let n be a positive integer. Let
A = Σₖ₌₀ⁿ ( - 1)ᵏⁿCₖ[ ( 1/2 )ᵏ + ( 3/4 )ᵏ + ( 7/8 )ᵏ + ( 15/16 )ᵏ + ( 31/32 )ᵏ ] . If
63A = 1 - 1/2³⁰, then n is equal to _____________.

Solution

Maths sir · 4 steps

Each bracket term is a binomial expansion in disguise: collapse each to (1-x)ⁿ, then sum a GP.

1. Collapse each sum. For any x, Σₖ₌₀ⁿ (-1)ᵏ ⁿCₖ xᵏ = (1-x)ⁿ. Splitting A into five such sums with x = 1/2, 3/4, 7/8, 15/16, 31/32 gives 1-x = 1/2, 1/4, 1/8, 1/16, 1/32.
2. Spot the GP. So A = 1/2ⁿ + 1/4ⁿ + 1/8ⁿ + 1/16ⁿ + 1/32ⁿ. With r = 1/2ⁿ this is r + r² + r³ + r⁴ + r⁵, a GP of 5 terms.

   A = r(1-r⁵)/(1-r) = 1/(2ⁿ - 1)(1 - 1/2⁵ⁿ)

3. Match with the data. Multiplying by 63 gives 63A = 63/(2ⁿ-1)(1 - 1/2⁵ⁿ). For this to equal 1 - 1/2³⁰, we need 2ⁿ - 1 = 63 and 5n = 30.
4. So. Both conditions give 2ⁿ = 64, that is, n = 6.

The idea to keep. Whenever you see (-1)ᵏ ⁿCₖ xᵏ summed over all k, write it straight away as (1-x)ⁿ. Then look for a pattern, often a GP, among the results.

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