A perpetuity pays 1 at the end of the 1st year, 2 at the end of the 2nd year, 3 at the end of the 3rd year, etc. Which of the following expressions give the present value of this perpetuity? I. . 1/i^2
II. 1−d/d^2
III. e^δ+e^−δ
A) I and II only
B) I and III only
C) II and III only
D) I, II and III

da freeck kinda math is this just drop out

step-by-step explanation:

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

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