If
Σₖ₌₁³¹ ( ³¹Cₖ )( ³¹Cₖ₋₁ ) - Σₖ₌₁³⁰ ( ³⁰Cₖ )( ³⁰Cₖ₋₁ ) = (α (60!))/(30!)(31!) ,
where α ∈ R, then the value of 16α is equal to

Solution

Maths sir · 5 steps

Turn ⁿCₖ₋₁ into ⁿCₙ₋ₖ₊₁, and each sum collapses to one coefficient by Vandermonde.

1. Flip one factor. Use ⁿCₖ₋₁ = ⁿCₙ₋ₖ₊₁. The two lower indices now add to k + (n-k+1) = n+1 in every term.

   Σₖ₌₁ⁿ ⁿCₖ ⁿCₙ₊₁₋ₖ = ²ⁿCₙ₊₁

2. Why it works. Each term is one way to pick xᵏ from one (1+x)ⁿ and xⁿ⁺¹⁻ᵏ from the other. So the sum is the coefficient of xⁿ⁺¹ in (1+x)²ⁿ. With n = 31 you get ⁶²C₃₂, and with n = 30 you get ⁶⁰C₃₁.
3. Common factor. ⁶²C₃₂ = 62!/(32! 30!) = (62 × 61)/32 · 60!/(31! 30!) = 1891/16 · 60!/(30! 31!). Also ⁶⁰C₃₁ = 60!/(31! 29!) = 30 · 60!/(30! 31!).
4. Find alpha. α = 1891/16 - 30 = (1891 - 480)/16 = 1411/16.
5. So. 16α = 1411, which is option (a).

The idea to keep. When a sum multiplies two binomial coefficients, make their lower indices add to a constant. Then it is one coefficient of (1+x)ᵐ⁺ⁿ by Vandermonde.

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