If Σᵣ₌₀¹⁰ ((10ʳ⁺¹ - 1)/10ʳ).¹¹Cᵣ₊₁ = (_α 11 - 11¹¹)/10¹⁰, then α is equal to:

Solution

Maths sir · 4 steps

Split the bracket into 10 - 1/10ʳ; then each half is a binomial sum you already know.

1. Split the term. Write (10ʳ⁺¹-1)/10ʳ = 10 - 1/10ʳ. The sum then breaks into two sums: 10Σ ¹¹Cᵣ₊₁ minus Σ ¹¹Cᵣ₊₁/10ʳ.
2. First sum. As r runs from 0 to 10, r+1 runs from 1 to 11. So Σ ¹¹Cᵣ₊₁ = 2¹¹ - ¹¹C₀ = 2¹¹ - 1, and the first part is 10(2¹¹-1).
3. Second sum. Put k = r+1, so 1/10ʳ = 10 × (1/10)ᵏ. Then Σₖ₌₁¹¹ ¹¹Cₖ × 10 (1/10)ᵏ = 10[(1+1/10)¹¹ - 1] = 10[(11/10)¹¹ - 1].

   10(2¹¹-1) - 10[11¹¹/10¹¹ - 1] = 10 × 2¹¹ - 11¹¹/10¹⁰ = (20¹¹ - 11¹¹)/10¹⁰

4. So. The two -10 and +10 terms cancel, and 10 × 2¹¹ = (10¹¹ × 2¹¹)/10¹⁰ = 20¹¹/10¹⁰. Matching with (α¹¹ - 11¹¹)/10¹⁰ gives α = 20, option (b).

The idea to keep. When a binomial sum has an awkward factor like 1/10ʳ, read it as a power of x and use (1+x)ⁿ. Watch the shifted index: the sum starts at ⁿC₁, so subtract the missing ⁿC₀ = 1.

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