Question 251550: At the beginning of each year for 5 years, Cindy invests in a mutual fund with an average annual return of ‘r’. The first year she invests $10, the second year $20, the third year $30, the forth year $40 and the fifth year $50. If we assume that x = 1 + r, then at the end of the first year (just before next investment) the value is 10x dollars. Because $20 is then added to the 10x dollars, the amount at the end of second year will be (10x + 20)x dollars and so on. Find:
a. A polynomial in ‘x’ that represents the value of the investment at the end of the fifth year.
b. How much money will Cindy have after 5 years if r = 8%?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! since the start of the second year coincides with the end of the first year, an investment at the start of the second year can also be viewed as an investment at the end of the first year.
end of first year she has 10x + 20
end of second year she has 10x^2 + 20x + 30
end of third year she has 10x^3 + 20x^2 + 30x + 40
end of fourth year she has 10x^4 + 20x^3 + 30x^2 + 40x + 50
end of fifth year she has 10x^5 + 20x^4 + 30x^3 + 40x^2 + 50x
the polynomial should be:
f = 10x^5 + 20x^4 + 30x^3 + 40x^2 + 50x
f = future value
if x = 1.08, this becomes:
f = 10*(1.08)^5 + 20*(1.08)^4 + 30*(1.08)^3 + 40*(1.08)^2 + 50*(1.08)
if we calculate the future value of her investment using this polynomial, then we would get:
f = 180.35042
we can duplicate this in real life to see if the answer is the same.
end of first year she has 10 * 1.08 = 10.8 + 20 = 30.8
end of second year she has 30.8 * 1.08 = 33.264 + 30 = 63.264
end of third year she has 63.264 * 1.08 = 68.32512 + 40 = 108.32512
end of fourth year she has 108.32512 * 1.08 = 116.9911296 + 50 = 166.9911296
end of fifth year she has 166.9911296 * 1.08 = 180.350423
since the answer is the same, we can assume that the polynomial is good.
the formula can be factored by 10 to get:
f = 10 * (1*(1.08)^5 + 2*(1.08)^4 + 3*(1.08)^3 + 4*(1.08)^2 + 5*(1.08))
the formula can be generalized into the following form:
 where:
f = future value
a = initial investment
n = number of time periods to invest
r = interest rate per time period
i = the individual investment period
example:
in our problem.
a = 10
n = 5
r = .08
when i = 1, the formula becomes:
10*1*(1.08)^5 = 10*(1.08)^5
when i = 2, the formula becomes:
10*2*(1.08)^4 = 20*(1.08)^4
when i = 3, the formula becomes:
10*3*(1.08)^3 = 30*(1.08)^3
when i = 4, the formula becomes:
10*4*(1.08)^2 = 40*(1.08)^2
when i = 5, the formula becomes:
10*5*(1.08)^1 = 50*(1.08)^1
putting all these together, you get:
f = 10*(1.08)^5 + 20*(1.08)^4 + 30*(1.08)^3 + 40*(1.08)^2 + 50*(1.08)
this is the same as the formula we initially derived except that it is now generated from a more general form of the same equation.
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