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Impact of the Compounding Frequency on the Growth of Investments

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Provided below is a formula that I already understand, to include finding the answer.

P=A(1/(1+r)^n ) = P=(5000)(1/(1+0.08)^12 )≌1985.60

The equation and variables below is similar to the above equation, However, the A is on the sum side. Can someone show me the steps in this process.

If the amount invested and the interest is compounded 8 times a year, please note that the formula is now A = P(1 + r/n)^nt, where A is the future balance, P is an initial balance, r is the interest rate, and n is the number of times that is compounded. To calculate this lets use t = 4 years and r = 8%, n = 8, and
A = $ 3000.

Understanding the method in solving will help me with many more like equations. Not sure if the variables and the arrangement of this equation have brain locked my thought process. I appreciate any assistance.

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Solution Summary

The solution uses few examples and quantitative calculations to show that the more frequent the interest is compounded the larger the initial invested amount grows.

Solution Preview

In the first formula we have the situation, when interest is compounded (that is credited to the account) annually. If P is the initial balance, then
after one year it will become P1 =P*(1+r) = P*(1+0.08) for r = 0.08 interest rate. Then, this amount is reinvested again with same interest rate 8 per cent.
Then, after the second year, the amount will be P2 = P1*(1+0.08) = P*(1+0.08)^2 Then, this amount P2 is re-invested at the end of the second year
to result in P3 = P2*(1+0.08) by the end of the third year. And so on, the procedure repeats 12 times. At the end of 12th year, the balance A will
be = P*(1+0.08)^12. So, we have A = P*(1+0.08)^12 Solving for P ...

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