If ²⁰Cᵣ is the co-efficient of xʳ in the expansion of (1 + x)²⁰, then the value of Σᵣ₌₀²⁰ r².²⁰Cᵣ is equal to :

Solution

Maths sir · 4 steps

Split r² as r(r-1) + r, because each of those pieces cancels neatly against the factorials in ²⁰Cᵣ.

1. Break up r squared. Write r² = r(r-1) + r. A bare r² does not simplify against ⁿCᵣ, but r(r-1) and r each do.
2. Use the factorial cancels. Use r · ⁿCᵣ = n · ⁿ⁻¹Cᵣ₋₁ and r(r-1) · ⁿCᵣ = n(n-1) · ⁿ⁻²Cᵣ₋₂. Summing over r turns each into a full row sum: Σ r · ⁿCᵣ = n · 2ⁿ⁻¹ and Σ r(r-1) · ⁿCᵣ = n(n-1) · 2ⁿ⁻².

   Σᵣ₌₀²⁰ r² ²⁰Cᵣ = 20 × 19 × 2¹⁸ + 20 × 2¹⁹

3. Combine. Write both terms with 2¹⁸. The first term is 380 × 2¹⁸, and the second is 20 × 2¹⁹ = 40 × 2¹⁸. Together they give 420 × 2¹⁸, which matches the formula n(n+1) 2ⁿ⁻² with n = 20.
4. So. The sum is 420 × 2¹⁸, option (d).

The idea to keep. For sums of r² · ⁿCᵣ, split r² = r(r-1) + r. The result is n(n+1) 2ⁿ⁻². Each extra factor of r pulls out one n and costs one power of 2.

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