MHB Compound Interest: $50000 Invested for 3 Years at 7.75%

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An alumnus of a local high school donated $50000 to the school. The amount was invested for 3 years at 7.75%, compounded quarterly. The school has agreed to use only the interest earned on the investment to buy sports equipment. How much money will be available for sports equipment at the end of the investments term?
 
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$A = A_0 \left(1 + \dfrac{r}{n}\right)^{nt}$

$A$ = account balance after $t$ years
$A_0$ = initial account balance
$r$ = annual interest rate as a decimal
$n$ = number of compounding periods per year
 
I tend to prefer not blindly substituting into a formula.

If the interest rate is 7.75% p.a. compounded quarterly, then the quarterly rate is 1.9375%.

So every quarter, you increase by 1.9375%, thus you end up with 101.9375%.

Thus the multiplier is 1.019375

If you're investing for 3 years, then that's 12 quarters.

Thus $\displaystyle A = 50\,000 \times \left(1.019\,375 \right)^{12} $.
 
Even less "blindly": The interest rate is 7.75% annually so 7.75/4= 1.9375% per quarter as Prove It said. That means that after 3 months (one quarter of a year) interest of 50000(0.019375) will be added to the 50000 yielding 50000+ 50000(0.019375)= 50000(1.019375). Since the interest is compounded (neither principle nor interest is collected) quarterly, the next quarter the 1.9375% interest will be calculated on that new 50000(1.019375) so the interest at the end of the second quarter will be (50000(1.019375))(0.019375) and that will be added to the 50000(1.019375) so will be (50000(1.019375))(0.019375)+ 50000(1.019375)= 50000(1.019375)(0.019375+ 1)= 50000(1.0199375)^2.

The third quarter you do the same thing except this time you start with 50000(1.01375)^2 so the interest will be 50000(1.019375)^2(0.019375) added to 50000(1.019375)^2 to get 50000(1.019375)^2(0.019375)+ 50000(1.019375)^2= 50000(1.019375)^2(0.019375+ 1)= 50000(1.019375)^3.

In 3 years there are 3(4)= 12 quarters so you repeat this 12 times getting Prove It's 50000(1.019375)^12.
 
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