Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7an−1−10an−2for n ≥ 2, a0= 2, a1= 1 c) an= 6an−1−8an−2for n ≥ 2, a0= 4, a1= 10 d) an= 2an−1−an−2for n ≥ 2, a0= 4, a1= 1 e) an= an−2for n ≥ 2, a0= 5, a1= -1 f) an=− 6an−1−9an−2for n ≥ 2, a0= 3, a1= -3 g) an+2 = -4an+15anfor n ≥ 0, a0= 2, a1= 8

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form , then it will satisfy ...

Rearranging and dividing by , we get the quadratic ...

The quadratic formula tells us values of r that satisfy this are ...

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

We can find p and q by solving the initial condition equations:

These have the solution ...

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Using these formulas on the first recurrence relation, we get ...

a)

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The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

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For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

The initial condition equations are now ...

and the solutions for p and q are ...

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Using these formulas on problem (d), we get ...

d)

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And for problem (f), we get ...

f)

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Comment on problem g

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

your answer is b ^w^ hope i

c

step-by-step explanation: