# hello new to consumer math need help

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• Mar 19th 2008, 07:03 AM
tdmj90
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
• Mar 19th 2008, 11:54 AM
iknowone
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 $400$,

after 2 years you have a balance of $400+400(1.04) = 400((1.04)^0+(1.04)^1)$

after 3 years your balance is $400+400(1.04)+400(1.04)^2 = 400((1.04)^0 + (1.04)^1+(1.04)^2)$

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 $1.04$ is a geometric series which simplifies to the formula: $\frac{1-(1.04)^n}{1-1.04}$

So after $n$ years you should have saved: $400\left(\frac{1-(1.04)^n}{1-1.04}\right)$.