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 ![a[n]=r^n](https://tex.z-dn.net/?f=a%5Bn%5D%3Dr%5En)
, 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 ...
![a[n]=pr_1^n+qr_2^n](https://tex.z-dn.net/?f=a%5Bn%5D%3Dpr_1%5En%2Bqr_2%5En)
We can find p and q by solving the initial condition equations:
![\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%261%5C%5Cr_1%26r_2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dp%5C%5Cq%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5B0%5D%5C%5Ca%5B1%5D%5Cend%7Barray%7D%5Cright%5D)
These have the solution ...
![p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7Ba%5B0%5Dr_2-a%5B1%5D%7D%7Br_2-r_1%7D%5C%5C%5C%5Cq%3D%5Cdfrac%7Ba%5B1%5D-a%5B0%5Dr_1%7D%7Br_2-r_1%7D)
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Using these formulas on the first recurrence relation, we get ...
a)
![c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n](https://tex.z-dn.net/?f=c_1%3D1%2C%5C%20c_2%3D6%2C%5C%20a%5B0%5D%3D3%2C%5C%20a%5B1%5D%3D6%5C%5C%5C%5Cr_1%3D%5Cdfrac%7B1%2B%5Csqrt%7B1%5E2%2B4%5Ccdot%206%7D%7D%7B2%7D%3D3%2C%5C%20r_2%3D%5Cdfrac%7B1-%5Csqrt%7B1%5E2%2B4%5Ccdot%206%7D%7D%7B2%7D%3D-2%5C%5C%5C%5Cp%3D%5Cdfrac%7B3%28-2%29-6%7D%7B-5%7D%3D%5Cdfrac%7B12%7D%7B5%7D%2C%5C%20q%3D%5Cdfrac%7B6-3%283%29%7D%7B-5%7D%3D%5Cdfrac%7B3%7D%7B5%7D%5C%5C%5C%5Ca%5Bn%5D%3D%5Cdfrac%7B3%7D%7B5%7D%28-2%29%5En%2B%5Cdfrac%7B12%7D%7B5%7D3%5En)
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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 ...
![a[n]=(p+qn)r^n](https://tex.z-dn.net/?f=a%5Bn%5D%3D%28p%2Bqn%29r%5En)
The initial condition equations are now ...
![\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5Cr%26r%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dp%5C%5Cq%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5B0%5D%5C%5Ca%5B1%5D%5Cend%7Barray%7D%5Cright%5D)
and the solutions for p and q are ...
![p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}](https://tex.z-dn.net/?f=p%3Da%5B0%5D%5C%5C%5C%5Cq%3D%5Cdfrac%7Ba%5B1%5D-a%5B0%5Dr%7D%7Br%7D)
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Using these formulas on problem (d), we get ...
d)
![c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n](https://tex.z-dn.net/?f=c_1%3D2%2C%5C%20c_2%3D-1%2C%5C%20a%5B0%5D%3D4%2C%5C%20a%5B1%5D%3D1%5C%5C%5C%5Cr%3D%5Cdfrac%7B2%2B%5Csqrt%7B2%5E2%2B4%28-1%29%7D%7D%7B2%7D%3D1%5C%5C%5C%5Cp%3D4%2C%5C%20q%3D%5Cdfrac%7B1-4%281%29%7D%7B1%7D%3D-3%5C%5C%5C%5Ca%5Bn%5D%3D4-3n)
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And for problem (f), we get ...
f)
![c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n](https://tex.z-dn.net/?f=c_1%3D-6%2C%5C%20c_2%3D-9%2C%5C%20a%5B0%5D%3D3%2C%5C%20a%5B1%5D%3D-3%5C%5C%5C%5Cr%3D%5Cdfrac%7B-6%2B%5Csqrt%7B6%5E2%2B4%28-9%29%7D%7D%7B2%7D%3D-3%5C%5C%5C%5Cp%3D3%2C%5C%20q%3D%5Cdfrac%7B-3-3%28-3%29%7D%7B-3%7D%3D-2%5C%5C%5C%5Ca%5Bn%5D%3D%283-2n%29%28-3%29%5En)
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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.
