The sum to the infinity of the series 1 + 2/3 + 6/3^2 + 10/3^3 + 14/3^4 + is: - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

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1 mark

The sum to the infinity of the series 1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots is:

Solution & Step-by-step Explanation

S = 1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \dots \frac{1}{3} S = \frac{1}{3} + \frac{2}{3 ^{2}} + \frac{6}{3 ^{3}} + \dots Subtracting: \frac{2}{3} S = 1 + \frac{1}{3} + \frac{4}{3 ^{2}} + \frac{4}{3 ^{3}} + \dots \frac{2}{3} S = \frac{4}{3} + [\frac{4}{3 ^{2}} + \frac{4}{3 ^{3}} + \dots] The term in bracket is a GP with a = 4/9 and r = 1/3 .Sum = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} = 2/3 . \frac{2}{3} S = \frac{4}{3} + \frac{2}{3} = 2 . S = 3 .

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The sum to the infinity of the series 1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots is:

2

3

4

6

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