Let m, n∈N and gcd (2, n) = 1. If 30( 30; 0 ) + 29( 30; 1 ) + ...... + 2( 30; 28 ) + 1( 30; 29 ) = n.2ᵐ, then n + m is equal to __________.
(Here ( n; k ) = ⁿCₖ)

Solution

Maths sir · 4 steps

Flip each coefficient with ³⁰Cₖ = ³⁰C₃₀₋ₖ, and the sum becomes a standard Σ r ⁿCᵣ.

1. Write the general term. The term with ³⁰Cₖ has multiplier 30-k, for k = 0 to 29. You can add the k = 30 term too, because its multiplier is 0. So the sum is Σₖ₌₀³⁰ (30-k) ³⁰Cₖ.
2. Use symmetry. Replace ³⁰Cₖ by ³⁰C₃₀₋ₖ and put r = 30-k. As k runs from 0 to 30, r also runs from 30 down to 0.

   Σₖ₌₀³⁰ (30-k) ³⁰Cₖ = Σᵣ₌₀³⁰ r ³⁰Cᵣ = 30 × 2²⁹

3. Make n odd. The condition gcd(2, n) = 1 means n must be odd. Since 30 = 15 × 2, we get 30 × 2²⁹ = 15 × 2³⁰, so n = 15 and m = 30.
4. So. n + m = 15 + 30 = 45.

The common slip: 59. Stopping at 30 × 2²⁹ and taking n = 30, m = 29 gives 59. But 30 is even, so gcd(2, n) = 1 fails. Move the extra 2 into the power first.

The idea to keep. When the multipliers run backwards against the coefficients, flip with ⁿCₖ = ⁿCₙ₋ₖ. Then use Σ r ⁿCᵣ = n 2ⁿ⁻¹. Always pull every factor of 2 into the power before you read off an odd n.

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