Solve the recurrence relation an= 2an-1 + 8an−2 + 81n2, a1 = −5, a2 = 202.

Given, recurrence relation-an=2an-1+8an-2+81n2 —(1)a1=-5a2=202(1)⇒an-2an-1-8an-2=81nCharacteristic equation is-rn2-2m-8=0=m2-4m+2m-8=0=(m-4)(m+2)=0=m1=-2, m2=4ah=C1(-2)n+C24n Let, f(n)=81n2f(n) is of the form An2+Bn+CTake at= An2+Bn+C(1)⇒ An2+Bn+C=2A(n-1)2+2B(n-1)+2C+8A(n-2)2+8B(n-2)+8C+81n2⇒An2+Bn+C=2A(n2+1-2n)+2Bn-2B+2C+8A(n2+4-4n)+8Bn-16B+8C+81n2⇒An2+Bn+C=n2(2A+8A+81)+n(-4Q+2B-32A+8B)+(2A-2B+2C+32A-16+8C) Comparing both sides-A=2A+8A+81 —(2)B=-4A+2B-32A+8B —(3)C=2A-2B+2C+32A-16B+8C —(4)(2)⇒9A=81⇒A=9(3)⇒9B=36A⇒B=4A⇒B=36(4)⇒9C=-34A+18B⇒9C=342⇒C=38so, at=9n2+36n+38Hence,an=C1(-2)n+C25n+9n2+36n+38 Again.a1=-5⇒-5=c1(-2)+c25+9+36+38⇒-2c1+5c2=-88 —(5)a2=202⇒202=4c1+25c2+36+72+38⇒4c1+25c2=56 —(6)solving (5) & (6) we get-c1=2487c2=-247an=2487(-2)n-247(5)n+9n2+36n+38

Answered: Solve the recurrence relation an= 2an-1…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.5: The Binomial Theorem

Problem 16E

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Solve the recurrence relation a_n= 2a_n-1 + 8a_n−2 + 81n²
, a₁ = −5, a₂ = 202.

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