The first terms of an infinite geometric sequence, un are 2, 6, 18, the first terms of a second infinite geometric sequence, vn are 2, -6, 18, - the terms of a third sequence wn are defined as wn = un + vn

1. the finite series, summation: upper limit= 225, lower limit k=1, formula wk
can also be written in the form summation: upper limit= m, lower limit k=0, formula 4r^k.
find the value of r.

2. find the value of m

r = 9 and m = 112

Step-by-step explanation:

Write W in terms of U and V.

Define U and V using geometric series formula.

Use sum of geometric series formula.

Simplify.

Therefore, r = 9 and m = 112.

in the attachment.   i didn't see your choices.   the word which means there is choices.

step-by-step explanation:

(x,y)

x tells you how much left or right you go from the origin

y tells you how much down or up you go from the origin

so for example (-3,2), you go left 3 and up twice. i denoted which of the ordered pairs is which in my graph.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt