Let $\left\{a_{\mathrm{n}}\right\}_{\mathrm{n}=0}^{\infty}$ be a sequence such that $a_{0}=a_{1}=0$ and $a_{\mathrm{n}+2}=3a_{\mathrm{n}+1}-2a_{\mathrm{n}}+1, \forall \mathrm{n} \geq 0$. Then $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to:

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Sequences and Series

Let

{a_{n}}_{n = 0}^{\infty}

be a sequence such that

a_{0} = a_{1} = 0

and

a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0

.
Then

a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}

is equal to:

Answer: (B)

From the recurrence

a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1

, we get

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

. Summing from

n = 0

n = n

gives

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

. Setting

n \to n - 2

, we have

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

. Therefore,

a_{25} - 2 a_{24} = 24

and

a_{23} - 2 a_{22} = 22

. The expression factors as

(a_{25} - 2 a_{24}) (a_{23} - 2 a_{22}) = (24) (22) = 528

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option