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The sum of the first n terms of an AP is Sₙ = 3n² - 2n for all natural numbers n. Find the value of the infinite sum: sumₙ₌₁^(infinity) of 21 / [(Sₙ * Sₙ₊₂ + Sₙ₋₁ * Sₙ₊₁) - (Sₙ * Sₙ₊₁ + Sₙ₋₁ * Sₙ₊₂)].
- 3
- 2
- 1
- 8
Correct answer: 2
Solution
The denominator simplifies to aₙ * aₙ₊₂ = (6n-5)(6n+7). Using partial fractions: 21/[(6n-5)(6n+7)] = (7/4)[1/(6n-5) - 1/(6n+7)]. The infinite sum telescopes in pairs (residuals at positions 1 and 7 survive) giving (7/4)(1 + 1/7) = (7/4)(8/7) = 2.
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