11. Show thatan=2+1 is a solution to the recurrence relationan= 2an-1-1 with a₁ =3.12. Consider 25 = 1 (mod 3) and 11 = 2 (mod 3),If 25* 11* = 4 mod 3, find x.13. Prove that the sum of the first n even positive integers is n(n-1)[0+2+4+6+....…..…….……………...]rove by mathematical induction that for every positive integer n,is divisible by 3.

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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4. Find an explicit formula for the recurrence relation an= An-1+6an-2 ao= 5 aj = 0 %3D
-9ал-2; ао — 1, ај — 2 Solve the recurrence relation an = 6an-1- 9an-2; ao =1, a1 = 2
Q2. Solve the recurrence relation an = ao = 1 and a, 7an-1 - 10 an-2 with initial conditions. = 2.
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7. Solve the recurrence relation A, = 4 An2- 3.An-1 subject to the initial conditions A, 5 and A, = 10.
13. Solve the recurrence relation. Given:a0=3 a1= 6 an= 6a n-1 + 7a n-2 Show your work in the space provided. a. Find c1 and c2 an= (6)a n-1 + (7)a n-2 c1= c2= b. Substitute c1 and c2 into the following equation: r^2- c1r - c2= 0 show your work here: c. Identify a, b, and c in the quadratic equation. a= b= c= d. Use the quadratic formula to find the two roots. Here is the quadratic formula: -b + Vb^2- 4ac/2a Show your work here: e. Substitute two roots, r1 and r2 into the equation an= a1r1^n+ a2r2^n Show your work here: f. Now substitute to find two equations, a0 and a1 Remember to use the equation you found from step e. show your work here: a0= 3 =a1= 6 = g. Add the two equations together find a1 and a2 a1= a2= h. What is the solution to the recurrence relations? an= i. Find the 10" term of the sequence, using the solution to the recurrence relation you just found. Show your work here: a10=
5. Solve the recurrence relation an = 3an-1- 2an-2 with initial conditions ao = a, a1 = b. Here a and b are constants. Зап-1 — 2ап-2 %3D
Solve the recurrence relation an = 7an-1 -12an-2 for all integers n ≥ 2, a = 1 and a₁ = 6.
Solve the following recurrence relations ( an = 5an-1 - 4an-2, ao = 0, a1 = 1 an = -10a,-1 - 21an-2, ao = 2, a1 = 1
What is the solution to the recurrence relation: Tn=5nTn-1+6 with T0=3?
tion 6 The value of the characteristics variable r in the characteristic ret equation of the recurrence relation an = -2an 1+3an 2 is ered ed out of Or=-1,-3 ag question Or= 1,3 Or= 1,-3 Or=-1,3 evious page Next pag

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how that an = 2n+1 is a solution to the recurrence relation an = 2an-1-1 with a₁ =3.