Answer:
- a) 3/5·((-2)^n + 4·3^n)
- b) 3·2^n - 5^n
- c) 3·2^n + 4^n
- d) 4 - 3 n
- e) 2 + 3·(-1)^n
- f) (-3)^n·(3 - 2n)
- g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19
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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<em>Comment on problem g</em>
Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.