Let the first term of a series be T₁=6 and its rᵗʰ term Tᵣ=3 Tᵣ₋₁+6ʳ, r=2,3, ............ n. If the sum of the first n terms of this series is 1/5(n²-12 n+39)(4 · 6ⁿ-5 · 3ⁿ+1), then n is equal to ___________.

Solution

Maths sir · 4 steps

Divide the recurrence by 3ʳ: it telescopes, and then you just match your sum with the given one.

1. Divide by 3^r. Dividing Tᵣ = 3Tᵣ₋₁ + 6ʳ by 3ʳ gives Tᵣ/3ʳ = Tᵣ₋₁/3ʳ⁻¹ + 2ʳ. Put Uᵣ = Tᵣ/3ʳ. Then U₁ = 6/3 = 2 and Uᵣ - Uᵣ₋₁ = 2ʳ.
2. Find the general term. Adding these differences gives Uᵣ = 2 + 2² + … + 2ʳ = 2ʳ⁺¹ - 2. So Tᵣ = 3ʳ(2ʳ⁺¹ - 2) = 2 · 6ʳ - 2 · 3ʳ. Check: T₂ = 72 - 18 = 54 = 3 × 6 + 36.
3. Add two GPs. Sₙ = 2Σ 6ʳ - 2Σ 3ʳ = 12(6ⁿ - 1)/5 - 3(3ⁿ - 1). Put both over 5: (12 · 6ⁿ - 12 - 15 · 3ⁿ + 15)/5.

   Sₙ = 3/5(4 · 6ⁿ - 5 · 3ⁿ + 1)

4. So. Matching with the given sum, n² - 12n + 39 = 3, that is n² - 12n + 36 = (n - 6)² = 0. Hence n = 6.

The idea to keep. For Tᵣ = aTᵣ₋₁ + bʳ, divide by aʳ. The recurrence becomes a plain difference you can add up, and Tᵣ comes out as a combination of two GPs.

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