For k ∈ N, let 1/(α (α + 1)(α + 2).........(α + 20)) = Σ_(K = 0)²⁰ Aₖ/(α + k) , where α > 0. Then the value of 100( (A₁₄ + A₁₅)/A₁₃ )² is equal to _____________.

Solution

Maths sir · 4 steps

Use the cover-up rule for each Aₖ, then divide them so the factorials mostly cancel.

1. Cover-up rule. To find Aₖ, multiply both sides by (α + k) and put α = -k. Every other term on the right becomes zero. So Aₖ is 1 divided by the product of all the other factors at α = -k.

   Aₖ = 1/Π_(j ≠ k)(j - k) = (-1)ᵏ/(k! (20-k)!)

2. Sign and factorials. The factors with j < k give (-1)(-2)…(-k) = (-1)ᵏ k!. The factors with j > k give 1 × 2 × … × (20-k) = (20-k)!.
3. Take ratios. A₁₄/A₁₃ = -(13! 7!)/(14! 6!) = -7/14 = -1/2 and A₁₅/A₁₃ = (13! 7!)/(15! 5!) = (7 × 6)/(15 × 14) = 1/5.
4. So. (A₁₄ + A₁₅)/A₁₃ = -1/2 + 1/5 = -3/10. Its square is 9/100, and 100 × 9/100 gives 9.

The common slip: 49. Dropping the (-1)ᵏ makes every Aₖ positive, so the bracket becomes 1/2 + 1/5 = 7/10 and the answer comes out as 49 instead of 9.

The idea to keep. For 1/α(α+1)…(α+n), the cover-up rule gives Aₖ = (-1)ᵏ/k!(n-k)!. Neighbouring Aₖ alternate in sign, so do not drop the (-1)ᵏ.

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