If ¹P₁ + 2.²P₂ + 3.³P₃ + .... + 15.¹⁵P₁₅ = ^qPᵣ - s,0 ≤ s ≤ 1, then ^(q + s)Cᵣ₋ₛ is equal to ______________.

Solution

Maths sir · 4 steps

Each term k · k! is (k+1)! - k! in disguise, so the whole sum telescopes.

1. Rewrite each term. ᵏPₖ = k!, so the general term is k · k!. Write k = (k+1) - 1, which gives k · k! = (k+1)! - k!.

   Σₖ₌₁¹⁵ k · k! = (2! - 1!) + (3! - 2!) + … + (16! - 15!) = 16! - 1

2. Match the form. The terms of the series are all of the form ⁿPₙ, so write 16! = ¹⁶P₁₆. Then the sum is ¹⁶P₁₆ - 1, giving q = 16, r = 16 and s = 1, which fits 0 ≤ s ≤ 1.
3. Evaluate. ^(q+s)Cᵣ₋ₛ = ¹⁷C₁₅ = ¹⁷C₂ = (17 × 16)/2 = 136.
4. So. The required value is 136.

The common slip: 680. Writing 16! as ¹⁶P₁₅ instead of ¹⁶P₁₆ gives ¹⁷C₁₄ = 680. Keep the ⁿPₙ pattern of the series.

The idea to keep. Whenever you see k · k! summed, split k as (k+1) - 1. The sum Σₖ₌₁ⁿ k · k! collapses to (n+1)! - 1.

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