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JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Sequences and Series

Consider the sequence

a_{1}, a_{2}, a_{3}, \dots

such that

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

and

a_{n + 2} = \frac{2}{a _{n + 1}} + a_{n}

for

n = 1, 2, 3, \dots

. If

(\frac{a _{1} + \frac{1}{a _{2}}}{a _{3}}) \cdot (\frac{a _{2} + \frac{1}{a _{3}}}{a _{4}}) \cdot (\frac{a _{3} + \frac{1}{a _{4}}}{a _{5}}) \dots (\frac{a _{30} + \frac{1}{a _{31}}}{a _{32}}) = 2^{α} (^{61} C_{31})

, then

α

is equal to :

(A)-30
(B)-31
(C)-60✓ Correct
(D)-61

Answer: (C)

Solution

From the recurrence

a_{n + 2} = \frac{2}{a _{n + 1}} + a_{n}

, we get

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

.

This means

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

forms an arithmetic progression:

2, 4, 6, 8, \dots

, so

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

.

Now:

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

The product becomes:

\prod_{n = 1}^{30} \frac{2 n + 1}{2 ( n + 1 )} = \frac{3 \cdot 5 \cdot 7 \dots 61}{2 ^{30} \cdot 2 \cdot 3 \dots 31} = \frac{61 ! / ( 2 ^{30} \cdot 30 !)}{2 ^{30} \cdot 31 !} = \frac{61 !}{2 ^{60} \cdot 31 ! \cdot 30 !} = \frac{1}{2 ^{60}} (31 61)

Therefore

α = - 60

.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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