Question 211671: 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! this is really 5 separate investments.
10 for 5 years
20 for 4 years
30 for 3 years
20 for 2 years
10 for 1 year
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at the end the sum of the investments will be $180.35042
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you can calculate it either way and you'll get the same answer.
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10 * 1.08 = 10.8
+ 20 = 30.8 * 1.08 = 33.264
+ 30 = 63.264 * 1.08 = 68.32512
+ 40 = 108.32512 * 1.08 = 116.9911296
+ 50 = 166.9911296 * 1.08 = 180.35042
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10*1.08^5 = 14.69328077
20*1.08^4 = 27.2097792
30*1.08^3 = 37.79136
40*1.08^2 = 46.656
50*1.08^1 = 54
sum = 180.35042
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best I could do was a polynomial that expressed the second relationship as follows:
n = 5
x = 10
i = .08
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y = f(x) =
x*(n-(n-1))*(1+i)^n +
x*(n-(n-2))*(1+i)^(n-1) +
x*(n-(n-3))*(1+i)^(n-2) +
x*(n-(n-4))*(1+i)^(n-3) +
x*(n-(n-5))*(1+i)^(n-4)
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This is how it works:
first term makes n-(n-1) = 5-4 = 1 and n = 5
we get:
10*(1)*(1.08)^5 = 10.8
second term makes n-(n-2) = 5-3 = 2 and n-1 = 4
we get:
10*(2)*(1.08)^4 = 27.2097792
third term makes n-(n-3) = 5-2 = 3 and n-2 = 3
we get:
10*(3)*(1.08)^3 = 37.79136
fourth term makes n-(n-4) = 5-1 = 4 and n-3 = 2
we get:
10*(4)*(1.08)^2 = 46.656
fifth term makes n-(n-5) = 5-0 = 5 and n-4 = 1
we get:
10*(5)*(1.08)^1 = 54
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sum of all of that is 180.35042 which is the same number that I had calculated earlier only this time it was derived from a polynomial with 3 factors.
Those are:
n = 5
x = 10
i = .08
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n determines the number of years and the number of terms to be used.
investment is made at the beginning of each year.
I tested it with n = 1, 2, 3, 4, 5, 6 and it works ok and provides results for the end of each year comparable to the first relationship I showed you where it showed the balance at the end of each year.
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