JEE Main 2022 · 25 June · evening · Single correct · +4 / −1
The sum 1 + 2 . 3 + 3 . 3² + ......... + 10 . 3⁹ is equal to :
- (a) (2 . 3¹² + 10)/4
- (b) (19 . 3¹⁰ + 1)/4
- (c) 5 . 3¹⁰ - 2
- (d) (9 . 3¹⁰ + 1)/2
Answer and solution
Answer (b) (19 . 3¹⁰ + 1)/4
Solution
Maths sir · 4 steps
Each term is (counting number) times (power of 3), so multiply by 3, shift and subtract.
- Spot the AGP. The general term is k · 3ᵏ⁻¹ for k = 1 to 10. The first factor is an AP and the second is a GP with ratio 3, so this is an arithmetico-geometric series. Call the sum S.
- Shift and subtract. Multiply by the ratio: 3S = 1 · 3 + 2 · 3² + … + 9 · 3⁹ + 10 · 3¹⁰. Write it one place to the right under S and subtract. Terms with the same power of 3 leave a coefficient of exactly 1.
S - 3S = (1 + 3 + 3² + … + 3⁹) - 10 · 3¹⁰
- Sum the GP. The bracket is a GP with 10 terms, so it equals (3¹⁰-1)/(3-1) = (3¹⁰-1)/2. Then -2S = (3¹⁰-1)/2 - 10 · 3¹⁰ = (-19 · 3¹⁰ - 1)/2.
- So. Dividing by -2 gives S = (19 · 3¹⁰ + 1)/4, which is option (b).
The idea to keep. For any AGP sum, multiply by the common ratio r, shift by one place and subtract. What remains is a plain GP minus one leftover last term.
Answer from the NTA final answer key. Solution by PrepWiser.