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JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Sequences and Series

Let

{a_{n}}_{n = 1}^{\infty}

be a sequence such that

a_{1} = 1, a_{2} = 1

and

a_{n + 2} = 2 a_{n + 1} + a_{n}

for all

n \geq 1

. Then the value of

47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}

is equal to........

Answer: 7

Solution

Let

A (x) = \sum_{n \geq 1} a_{n} x^{n}

. Using

a_{1} = a_{2} = 1

and

a_{n + 2} = 2 a_{n + 1} + a_{n}

,

A (x) = x + x^{2} + n \geq 1 \sum (2 a_{n + 1} + a_{n}) x^{n + 2} = x + x^{2} + 2 x (A (x) - x) + x^{2} A (x)

so

A (x) (1 - 2 x - x^{2}) = x - x^{2}

and

A (x) = \frac{x - x ^{2}}{1 - 2 x - x ^{2}}

. The required sum is

A (\frac{1}{8})

, and

x - x^{2} = \frac{1}{8} - \frac{1}{64} = \frac{7}{64}, 1 - 2 x - x^{2} = 1 - \frac{1}{4} - \frac{1}{64} = \frac{47}{64}

so

A (\frac{1}{8}) = \frac{7}{47}

and

47 n = 1 \sum \infty \frac{a _{n}}{2 ^{3 n}} = 7

.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer

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