Consider the sequence a₁, a₂, a₃, … such that a₁=1, a₂=2 and aₙ₊₂=2/aₙ₊₁+aₙ for n=1,2,3, … . If ((a₁+1/a₂)/a₃) ·((a₂+1/a₃)/a₄) ·((a₃+1/a₄)/a₅) …((a₃₀+1/a₃₁)/a₃₂)=2^α( ⁶¹ C₃₁), then α is equal to :

Solution

Maths sir · 4 steps

Multiply the recurrence by aₙ₊₁ and the products aₙaₙ₊₁ go up by exactly 2 each time.

1. Clear the fraction. Multiply aₙ₊₂ = 2/aₙ₊₁ + aₙ by aₙ₊₁. This gives aₙ₊₁aₙ₊₂ = aₙaₙ₊₁ + 2. So bₙ = aₙaₙ₊₁ is an AP with b₁ = 1 × 2 = 2 and common difference 2, which means bₙ = 2n.
2. Rewrite each factor. Multiply the top and bottom of the nth factor by aₙ₊₁. The top becomes aₙaₙ₊₁ + 1 = 2n + 1 and the bottom becomes aₙ₊₁aₙ₊₂ = 2n + 2. Check with n = 1: a₃ = 2, and (1 + 1/2)/2 = 3/4.

   Πₙ₌₁³⁰(2n+1)/(2n+2) = (3 · 5 … 61)/(4 · 6 … 62)

3. Turn into factorials. The odd product is 1 · 3 … 61 = 61!/(2 · 4 … 60) = 61!/(2³⁰ · 30!). The even product is 4 · 6 … 62 = 2³⁰(2 · 3 … 31) = 2³⁰ · 31!.
4. So. The product is 61!/(2⁶⁰ · 30! · 31!) = 2⁻⁶⁰ ⁶¹C₃₁, so α = -60, option (c).

The idea to keep. If a recurrence has 1/aₙ₊₁ in it, multiply through by aₙ₊₁. The product aₙaₙ₊₁ is often a simple AP. Write a product of odd numbers as a factorial over 2ᵏ · k!.

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