Question 1038459
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The amount next year is always the amount this year plus 6% times the amount this year.


So lets look at what happens the first couple of years. Since t is the number of years after 1997, in 1997, t = 0, in 1998, t = 1, and so on.

<pre>
   t     Previous       Amount Added           Total
-------------------------------------------------------------------------------------------------------------------
   0      0             43.2                    0    +  43.2
   1      43.2          43.2(0.06)             43.2  +  43.2(0.06) = 43.2(1.06)
   2      43.2(1.06)    [43.2(1.06)](0.06)     43.2(1.06) + 43.2(1.06)(0.06) = 43.2(1.06)(1.06) = 43.2(1.06)²

</pre>

I'll leave it to you to verify that for *[tex \Large t\ =\ 3] the total is *[tex \Large 43.2(1.06)^3].


So, if we use the facts that *[tex \Large a^0\ =\ 1] and *[tex \Large a^1\ =\ a] for any real numbers *[tex \Large a], then we can clean up our chart thus:

<pre>
   t        Total
-----------------------------
   0        43.2(1.06)^0
   1        43.2(1.06)^1
   2        43.2(1.06)^2
   3        43.2(1.06)^3
        .
        .
        .
   n        43.2(1.06)^n

</pre>


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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