Let {an}n=0∞ be a sequence such that a0 = a1 = 0 and an+2 = 3an+1 – 2an + 1, ∀n≥0. . Then a25a23−2a25a22−2a23a24+4a22a24 is equal to

a0 = 0, a1 = 0 an+2 = 3an+1 – 2an + 1 a25 a23 – 2a25a22 – a23a24 + 4a22a24 =? a2 = 3a1 – 2a0 + 1 a3 = 3a2 – 2a1 + 1 a4 = 3a3 – 2a2 + 1 a5 = 3a4 – 2a3 + 1 an+2 = 3an+1 – 2an + 1 ( + ) ⇒ ( a 2 + a 3 + a 4 + . . . . + a n + 1 + a n + 2 ) ⇒ an+2 = 2 (a2 + a3 + …. + an + an+1) –2 (a1 + a2 + ….+ an) + n + 1 an+2 = 2an+1 + n + 1 a25 a23 -2a25 a22 -a23 a24 + 4a22 a24 = a25 (a23 – 2a22) -2a24 (a23 – 2a22) As an+2 = 2an+1 + n + 1 ⇒ an+2 – 2an+1 = n + 1 ⇒ an+1 -2an = n ⇒ 24 * 22 = 528

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a₀ = a₁ = 0 and a_n+2 = 3_an+1 – 2a_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

Option 1 - 483

Option 2 - 528

Option 3 - 575

Option 4 - 624

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Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 2

Detailed Solution:

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2 = 3a_n+1 – 2a_n + 1

$(+) \Rightarrow (a_{2} + a_{3} + a_{4} + . . . . + a_{n + 1} + a_{n + 2})$

⇒ a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃

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