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

2 answers

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