If aᵣ is the coefficient of x¹⁰⁻ʳ in the Binomial expansion of (1 + x)¹⁰, then Σᵣ₌₁¹⁰ r³( aᵣ/aᵣ₋₁ )² is equal to

Solution

Maths sir · 5 steps

Turn the neighbour ratio into (11-r)/r and the sum becomes a plain polynomial sum.

1. Identify the coefficient. The coefficient of x¹⁰⁻ʳ in (1+x)¹⁰ is C(10, 10-r) = C(10, r), so aᵣ = C(10, r).
2. Ratio of neighbours. For consecutive binomial coefficients, C(n, r)/C(n, r-1) = (n-r+1)/r. With n = 10 this gives aᵣ/aᵣ₋₁ = (11-r)/r, so each term of the sum simplifies as below.

   r³((11-r)/r)² = r(11-r)²

3. Flip the index. Put s = 11-r. As r runs from 1 to 10, s runs from 10 down to 1. Each term becomes (11-s)s² = 11s² - s³, so the sum is 11Σ s² - Σ s³ over s = 1 to 10.
4. Standard sums. Σₛ₌₁¹⁰ s² = (10 × 11 × 21)/6 = 385 and Σₛ₌₁¹⁰ s³ = ((10 × 11)/2)² = 3025.
5. So. The sum is 11 × 385 - 3025 = 4235 - 3025 = 1210, which is option (d).

The idea to keep. Whenever a ratio of consecutive binomial coefficients appears, replace it with (n-r+1)/r at once. The sum usually collapses into Σ r² and Σ r³.

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