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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Sequences and Series

Let

a_{1} = b_{1} = 1

and

a_{n} = a_{n - 1} + (n - 1)

,

b_{n} = b_{n - 1} + a_{n - 1}

,

\forall n \geq 2

. If

S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}

and

T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}

, then

2^{7} (2 S - T)

is equal to

Answer: 461

Solution

Summing

a_{n} - a_{n - 1} = n - 1

gives

a_{n} = 1 + \frac{( n - 1 ) n}{2}

, while

b_{n} - b_{n - 1} = a_{n - 1}

produces

b_{1}, \dots, b_{10} = 1, 2, 4, 8, 15, 26, 42, 64, 93, 130

. Then

S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}} = \frac{3852}{1024}

, so

2 S = \frac{963}{128}

, and

T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}} = \frac{502}{128} = \frac{251}{64}

. Thus

2 S - T = \frac{963}{128} - \frac{502}{128} = \frac{461}{128}

, and

2^{7} (2 S - T) = 128 \cdot \frac{461}{128} = 461

.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer

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