Let <math><mrow><msubsup><mrow><mrow><mo>{</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msub></mrow><mo>}</mo></mrow></mrow><mrow><mi>n</mi><mo>=</mo><mn>0</mn></mrow><mrow><mo>∞</mo></mrow></msubsup></mrow></math> be a sequence such that a0 = a1 = 0 and an+2 = 3an+1 – 2an + 1, <math><mo>∀</mo><mi>n</mi><mo>≥</mo><mn>0</mn>. </math>. Then <math><mrow><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>5</mn></mrow></msub><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>3</mn></mrow></msub><mo>−</mo><mn>2</mn><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>5</mn></mrow></msub><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>2</mn></mrow></msub><mo>−</mo><mn>2</mn><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>3</mn></mrow></msub><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>4</mn></mrow></msub><mo>+</mo><mn>4</mn><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>2</mn></mrow></msub><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn><mn>4</mn></mrow></msub></mrow></math> is equal to

a0 = 0, a1 = 0an+2 = 3an+1 – 2an + 1a25 a23 – 2a25a22 – a23a24 + 4a22a24 =? a2 = 3a1 – 2a0 + 1a3 = 3a2 – 2a1 + 1a4 = 3a3 – 2a2 + 1a5 = 3a4 – 2a3 + 1an+2 = 3an+1 – 2an + 1 ( + ) ⇒ ( a 2 + a 3 + a 4 + . . . . + a n + 1 + a n + 2 ) <!- [if gte mso 9]><xml> </xml><! [endif]-> ⇒ an+2 = 2 (a2 + a3 + …. + an + an+1) –2 (a1 + a2 + ….+ an) + n + 1an+2 = 2an+1 + n + 1a25 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

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Maths Class 11th Relations and Functions

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

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 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 a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

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

⇒ a_n+2 – 2a_n+1 = n + 1

⇒ a_n+1 -2a_n = n

⇒ 24 × 22 = 528

a0 = 0, a1 = 0an+2 = 3an+1 – 2an + 1a25 a23 – 2a25a22 – a23a24 + 4a22a24 =? a2 = 3a1 – 2a0 + 1a3 = 3a2 – 2a1 + 1a4 = 3a3 – 2a2 + 1a5 = 3a4 – 2a3 + 1an+2  = 3an+1 – 2an + 1 <math> <mrow> <mrow> <mo> (</mo> <mrow> <mo>+</mo> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> <mo>+</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>a</mi> </mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815543014"> </o:OLEObject></xml><! [endif]-> ⇒ an+2 = 2 (a2 + a3 + …. + an + an+1) –2 (a1 + a2 + ….+ an) + n + 1an+2 = 2an+1 + n + 1a25 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

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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

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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

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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