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

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- Mar 19th 2008, 07:03 AMtdmj90hello 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 - Mar 19th 2008, 11:54 AMiknowone
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: $\displaystyle 400\left(\frac{1-(1.04)^n}{1-1.04}\right)$.