5 times the sum of 7 and 23

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= 7a

8_murik_8 [283]

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 $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

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

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

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 ...

$a[n]=pr_1^n+qr_2^n$

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

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

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

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

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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.)

_____

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 ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

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

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

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

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

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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.

6 0

2 years ago

Aleena expands 2xx+3-5(x-2) as 2x2+3-5x-10. Has she expanded it correctly? Please identify the mistakes and write the correct wa

Nimfa-mama [501]

The given expression is :

2xx+3-5(x-2)

We know that, $x{\cdot}x=x^2$

$=2x^2+3-5(x-2)\\\\=2x^2+3+(-5)(x)+(-5)(-2)\\\\=2x^2+3-5x+10$

Aleena expands 2xx+3-5(x-2) as 2x²+3-5x-10. She is mistaken because the sign before 10 should be +10 instead of -10. Hence, the correct expanded form is 2x²+3-5x+10. Hence, this is the required solution.

5 0

2 years ago

(pls help my teacher wants me to pass this!) which is the graph of this line segment after it is translated - 5 units along the

Kryger [21]

Answer:

3-th picture

Step-by-step explanation:

i think 3-th picture)

5 0

1 year ago

Read 2 more answers

The swim club rents the municipal pool for a flat fee of $500 plus a charge of $75 per swimmer during

xxTIMURxx [149]

Answer:

Y=500x+75

Step-by-step explanation:

4 0

3 years ago

The area of a rectangle is expressed as x4 - 9y2. What are the possible dimensions of the rectangle?

AnnyKZ [126]

This is a stupid question. The teacher who asked this needs to go back to school.

A = x⁴ - 9y²

That's going to be a number. As long as it's positive we can choose any length for our rectangle L and we can compute the width by W=A/L. So the correct answer is the dimensions are L by (x⁴ - 9y²)/L.

They don't want the correct answer. They want you to factor this expression. We'll do it, but don't believe for a second this factoring somehow constrains the possible dimensions of some rectangle.

We use the difference of two squares, a²-b² = (a+b)(a-b)

A = x⁴ - 9y² = (x²)² - (3y)² = (x²+3y)(x²-3y)

Answer: x²+3y by x²-3y

5 0

3 years ago