Sequence and Series
Arithmetico-Geometric Progression
JEE Main 2022
Grade 11

Question:

Let $S = 2 + \frac{6}{7} + \frac{12}{7^2} + \frac{20}{7^3} + \frac{30}{7^4} + \dots$ Then $4S$ is equal to:
(1) $\frac{7^3}{3^3}$
(2) $\frac{3^3}{7^3}$
(3) $\frac{343}{27}$
(4) $\frac{27}{343}$

Step-by-Step Solution

Key Concept: The numerators $2, 6, 12, 20, \dots$ are $n(n + 1)$. So $S = \sum_{n=1}^\infty n(n + 1)x^{n-1}$ with $x = 1/7$. Use the identity $\sum_{n=1}^\infty n(n + 1)x^{n-1} = 2/(1 - x)^3$.
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Correct Answer: (3)

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