The value of
-¹⁵C₁ + 2.¹⁵C₂ – 3.¹⁵C₃ + ... - 15.¹⁵C₁₅ + ¹⁴C₁ + ¹⁴C₃ + ¹⁴C₅ + ...+ ¹⁴C₁₁ is :

Solution

Maths sir · 3 steps

Pull out r with r ⁿCᵣ = n ⁿ⁻¹Cᵣ₋₁, then notice the odd series stops one term early.

1. First part. The term with ¹⁵Cᵣ is (-1)ʳ r ¹⁵Cᵣ for r = 1 to 15. Use r ¹⁵Cᵣ = 15 ¹⁴Cᵣ₋₁ to remove the factor r.

   Σᵣ₌₁¹⁵ (-1)ʳ r ¹⁵Cᵣ = -15Σₛ₌₀¹⁴ (-1)ˢ ¹⁴Cₛ = -15 (1-1)¹⁴ = 0

2. Second part. The full sum of odd-indexed coefficients is ¹⁴C₁ + ¹⁴C₃ + … + ¹⁴C₁₃ = 2¹³. The given series stops at ¹⁴C₁₁, so ¹⁴C₁₃ = ¹⁴C₁ = 14 is missing.
3. So. The value is 0 + (2¹³ - 14) = 2¹³ - 14, which is option (d).

The idea to keep. When you see r ⁿCᵣ, rewrite it as n ⁿ⁻¹Cᵣ₋₁ straight away. And always check the last term of a series: a missing term changes the answer.

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