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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Sequences and Series

If

S = \frac{7}{5} + \frac{9}{5 ^{2}} + \frac{13}{5 ^{3}} + \frac{19}{5 ^{4}} + \dots .

. , then 160 S is equal to ____.

Answer: 305

Solution

The numerators

7, 9, 13, 19, \dots

have constant second differences, with general term

n^{2} - n + 7

, so

S = \sum_{n \geq 1} \frac{n ^{2} - n + 7}{5 ^{n}}

. Using

\sum_{n \geq 1} \frac{1}{5 ^{n}} = \frac{1}{4}

,

\sum_{n \geq 1} \frac{n}{5 ^{n}} = \frac{5}{16}

and

\sum_{n \geq 1} \frac{n ^{2}}{5 ^{n}} = \frac{15}{32}

,

S = \frac{15}{32} - \frac{5}{16} + 7 \cdot \frac{1}{4} = \frac{61}{32}

. Therefore

160 S = 160 \cdot \frac{61}{32} = 305

.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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