Question 676179
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P(1\ +\ r)^t]


Is the accumulated value of *[tex \LARGE P] at *[tex \LARGE r] per annum expressed as a decimal for *[tex \LARGE t] years compounded annually, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 400(1\ +\ 0.05)^3]


The rest is just a bit of calculator work.  *[tex \LARGE A] is then the total amount you have to pay back and represents the original principal PLUS the interest.  So if you only want to know the amount of the interest, subtract *[tex \LARGE P] from *[tex \LARGE A].


By the way, if you compound more frequently than annually,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P\left(1\ +\ \frac{r}{n}\right)^{nt}]


Where *[tex \LARGE n] is the number of compounding periods per year.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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