The sum 1+3+11+25+45+71+… upto 20 terms, is equal to

Solution

Maths sir · 4 steps

The gaps between terms form an AP, so the nth term is a quadratic in n. Find it, then sum it.

1. Look at differences. The gaps between consecutive terms are 2, 8, 14, 20, 26, …, an AP with common difference 6. Constant second differences mean Tₙ is a quadratic in n.
2. Find the nth term. Start at 1 and add the first n-1 gaps: Tₙ = 1 + (n-1)/2[2 × 2 + (n-2) × 6] = 1 + (n-1)(3n-4). This gives Tₙ = 3n² - 7n + 5. Check: T₃ = 27 - 21 + 5 = 11 and T₄ = 48 - 28 + 5 = 25.

   S₂₀ = 3Σₙ₌₁²⁰ n² - 7Σₙ₌₁²⁰ n + 5 × 20

3. Put in numbers. Σ n² = (20 × 21 × 41)/6 = 2870 and Σ n = (20 × 21)/2 = 210. So S₂₀ = 8610 - 1470 + 100.
4. So. S₂₀ = 7240, which is option (a).

The idea to keep. When the terms do not form an AP or GP, take differences. If the differences form an AP, Tₙ is a quadratic in n, and you sum it with Σ n², Σ n and Σ 1.

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