Sue invested $1000 at 6% per year compounded yearly.Find the values of joe's investment at the end of each of the first five years
after the nth year, it is worth
1000(1+.06)^n
Why are you calling this trig ?
assuming his first deposit is NOW
end of 1st year: 1000(1.06) + 1000 = 2060
end of 2nd year: 2060(1.06) + 1000 = 3183.60
...
end of 5th year: .......... = 6637.09
fill in the rest using your calculator
To find the value of Joe's investment at the end of each of the first five years, we will use the compound interest formula:
A = P(1 + r/n)^(n*t)
Where:
A = the final amount (value of investment)
P = the principal amount (initial investment)
r = the annual interest rate (in decimal form)
n = the number of times that interest is compounded per year
t = the number of years
In this case, Sue's initial investment is $1000, the interest rate is 6% (or 0.06 as a decimal), and the interest is compounded yearly (n = 1).
Let's calculate the value of Sue's investment at the end of each of the first five years using the compound interest formula:
Year 1:
A = 1000(1 + 0.06/1)^(1*1)
A = 1000(1 + 0.06)^1
A = 1000(1.06)
A = $1060
Year 2:
A = 1000(1 + 0.06/1)^(1*2)
A = 1000(1 + 0.06)^2
A = 1000(1.06^2)
A = $1123.60
Year 3:
A = 1000(1 + 0.06/1)^(1*3)
A = 1000(1 + 0.06)^3
A = 1000(1.06^3)
A = $1191.02
Year 4:
A = 1000(1 + 0.06/1)^(1*4)
A = 1000(1 + 0.06)^4
A = 1000(1.06^4)
A = $1263.85
Year 5:
A = 1000(1 + 0.06/1)^(1*5)
A = 1000(1 + 0.06)^5
A = 1000(1.06^5)
A = $1341.96
Therefore, the values of Joe's investment at the end of each of the first five years are as follows:
Year 1: $1060
Year 2: $1123.60
Year 3: $1191.02
Year 4: $1263.85
Year 5: $1341.96