Intermediate Algebra Question!?

13.

a)

8-3 = 5

13-8 = 5

so slope = 5

a(n) = 5n + ?

3 = 5(1) + ?

? = -2

a(n) = 5n - 2

b)

1/16 - 1/4 = -3/16

1/64 - 1/16 = -3/64

this is a geometric sequence

1/16 / 1/4 = 1/4

1/64 / 1/16 = 1/4

so r=1/4

a(n) = a1(r)^(n-1)

1/4 = 1/4(1/4)^(1-1) = 1/4

so a(n) = 1/4(1/4)^(n-1)

15.

a)

summation notation can't be written here. a picture of it is here:

http://www.google.com/imgres?um=1&hl=en&sa=N&biw=1...

going from there, you first must find the equation:

19-23=-4

15-19=-4

so slope = -4

23 = -4(1) + ?

? = 27

a(n) = -4n + 27

stopping point:

3 = -4n + 27

n = 6

typical element: (-4n + 27)

starting point: 1

b)

arithmetic series?

-8 - 1 = -9

27 - (-8) = 35

geometric series?

-8/1 = -8

27/(-8) = -3.375

this is a special series.

we know that 1 = 1^3

8 = 2^3

27 = 3^3

64 = 4^3

125 = 5^3

and it alternates between positive and negative, starting with positive

so,

the equation is: a(n) = (-1)^(n-1) * n^3 (there are multiple ways to write this, but ill stick w/ this since it shows you how i got it)

stopping point:

125 = n^3

n = 5

typical element: ((-1)^(n-1) * n^3)

starting point: 1

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