JEE Main 2022 · 29 June · evening · Single correct · +4 / −1
The sum of the infinite series 1 + 5/6 + 12/6² + 22/6³ + 35/6⁴ + 51/6⁵ + 70/6⁶ + ..... is equal to :
- (a) 425/216
- (b) 429/216
- (c) 288/125
- (d) 280/125
Answer and solution
Answer (c) 288/125
Solution
Maths sir · 5 steps
The numerators have constant second differences, so multiply by (1-x) twice and a plain GP appears.
- Spot the pattern. The numerators are 1, 5, 12, 22, 35, 51, 70. Their differences are 4, 7, 10, 13, 16, 19, an AP with common difference 3. Put x = 1/6, so the series is S = 1 + 5x + 12x² + 22x³ + …
- First subtraction. Multiply by (1-x), which subtracts each numerator's neighbour. This gives (1-x)S = 1 + 4x + 7x² + 10x³ + …, so the new numerators are 1 followed by the differences.
- Second subtraction. Multiply by (1-x) again. Now (1-x)² S = 1 + 3x + 3x² + 3x³ + …, and after the first term it is a GP with ratio x.
(1-x)² S = 1 + 3x/(1-x)
- Substitute x. With x = 1/6, 1 - x = 5/6 and 3x/(1-x) = (1/2)/(5/6) = 3/5. So 25/36S = 1 + 3/5 = 8/5.
- So. S = 8/5 × 36/25 = 288/125, which is option (c).
The idea to keep. If the numerators become constant after k rounds of differences, multiply by (1-x) that many times. Each round peels off one level until a plain GP is left to sum.
Answer from the NTA final answer key. Solution by PrepWiser.