# Thread: hello new to consumer math need help

1. ## hello new to consumer math need help

How much can you accumulate in a savings plan if your annual year-end deposit is $400, and you earn 4% annual interest for 12 years? S =[(P(1+r)n – 1]/r can someone help simplify this for me? thank you 2. Once way you can do this is to set a table: [HTML]<table border = ".1" cellpadding = "4"> <tr> <td> </td><td> Deposited</td><td>Interest Earned</td><td>Total Balance</td> </tr> <tr> <td>End of year 1</td><td>400</td><td>0</td><td>400</td> </tr> <tr> <td>End of year 2</td><td>400</td><td>.04*400 = 16</td><td>400+400+16 = 816</td> </tr> <tr> <td>End of year 3</td><td>400</td><td>.04*816 = 32.64</td><td>816+400+32.64 = 1248.64</td> </tr> </table>[/HTML] and continue in this fashion for 12 years. Or observe that: after 1 year you have a balance of$\displaystyle 400$, after 2 years you have a balance of$\displaystyle 400+400(1.04) = 400((1.04)^0+(1.04)^1)$after 3 years your balance is$\displaystyle 400+400(1.04)+400(1.04)^2 = 400((1.04)^0 + (1.04)^1+(1.04)^2)$and after n years you have a balance of 400((1.04)^0 + (1.04)^1 + ... + (1.04)^{n-1}) The sum of powers of$\displaystyle 1.04$is a geometric series which simplifies to the formula:$\displaystyle \frac{1-(1.04)^n}{1-1.04}$So after$\displaystyle n$years you should have saved:$\displaystyle 400\left(\frac{1-(1.04)^n}{1-1.04}\right)\$.