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JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Sequences and Series

Let

α

and

β

be two real numbers such that

α + β = 1

and

α β = - 1

. Let

P_{n} = (α)^{n} + (β)^{n}, P_{n - 1} = 11

and

P_{n + 1} = 29

for some integer

n \geq 1

. Then, the value of

p_{n}^{2}

is ______

Answer: 324

Solution

The recurrence relation for

P_{n}

is

P_{n} = (α + β) P_{n - 1} - α β P_{n - 2} = 1 \cdot P_{n - 1} - (- 1) P_{n - 2} = P_{n - 1} + P_{n - 2}

. Given

P_{n - 1} = 11

and

P_{n + 1} = 29

. From the recurrence:

P_{n + 1} = P_{n} + P_{n - 1}

, so

29 = P_{n} + 11

, giving

P_{n} = 18

. Therefore

P_{n}^{2} = 1 8^{2} = 324

.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Numerical answer

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