JEE Main 2022 · 29 July · evening · Single correct · +4 / −1

Let {aₙ}ₙ₌₀^∞ be a sequence such that a₀=a₁=0 and ; ; aₙ₊₂=3 aₙ₊₁-2 aₙ+1, ∀ n ≥ 0 .
Then a₂₅ a₂₃-2 a₂₅ a₂₂-2 a₂₃ a₂₄+4 a₂₂ a₂₄ is equal to

  1. (a) 483
  2. (b) 528
  3. (c) 575
  4. (d) 624
Answer and solution

Answer (b) 528

Solution

Maths sir · 4 steps

Factor the expression first, then notice that aₙ₊₁-2aₙ goes up by exactly 1 each step.

  1. Factor by grouping. Take a₂₅ out of the first two terms and -2a₂₄ out of the last two. Both groups leave the same bracket (a₂₃-2a₂₂).

    a₂₅a₂₃-2a₂₅a₂₂-2a₂₃a₂₄+4a₂₂a₂₄ = (a₂₅-2a₂₄)(a₂₃-2a₂₂)

  2. Rewrite the recurrence. Move 2aₙ₊₁ to the left: aₙ₊₂-2aₙ₊₁ = aₙ₊₁-2aₙ+1. So if cₙ = aₙ₊₁-2aₙ, then cₙ₊₁ = cₙ + 1.
  3. Find the pattern. c₀ = a₁ - 2a₀ = 0, and each step adds 1, so cₙ = n. Hence a₂₅-2a₂₄ = c₂₄ = 24 and a₂₃-2a₂₂ = c₂₂ = 22.
  4. So. The expression equals 24 × 22 = 528, which is option (b).

The idea to keep. When a messy product of sequence terms appears, try grouping it into brackets like aₙ₊₁-kaₙ. Then rearrange the recurrence so it says how that same bracket changes.

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

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Asked about 250 times in JEE Main since 2020. This question is from 2022.