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JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Sequences and Series

If the sum of the series

(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2 ^{2}} - \frac{1}{2.3} + \frac{1}{3 ^{2}}) + (\frac{1}{2 ^{3}} - \frac{1}{2 ^{2} .3} + \frac{1}{2. 3 ^{2}} - \frac{1}{3 ^{3}}) + (\frac{1}{2 ^{4}} - \frac{1}{2 ^{3} \cdot 3} + \frac{1}{2 ^{2} \cdot 3 ^{2}} - \frac{1}{2 \cdot 3 ^{3}} + \frac{1}{3 ^{4}}) + \dots

is

\frac{α}{β}

, where

α

and

β

are co-prime, then

α + 3 β

is equal to

Answer: 7

Solution

The

n

th bracket is

j = 0 \sum n (\frac{1}{2})^{n - j} (- \frac{1}{3})^{j} = \frac{( \frac{1}{2} ) ^{n + 1} - ( - \frac{1}{3} ) ^{n + 1}}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} [(\frac{1}{2})^{n + 1} - (- \frac{1}{3})^{n + 1}]

. Summing over

n \geq 1

gives two geometric series,

\sum_{n \geq 1} (\frac{1}{2})^{n + 1} = \frac{1}{2}

and

\sum_{n \geq 1} (- \frac{1}{3})^{n + 1} = \frac{1}{12}

, so the sum is

\frac{6}{5} (\frac{1}{2} - \frac{1}{12}) = \frac{6}{5} \cdot \frac{5}{12} = \frac{1}{2}

. Hence

α = 1, β = 2

and

α + 3 β = 1 + 6 = 7

.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer

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