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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.

https://brainmass.com/math/basic-algebra/impact-compounding-frequency-growth-investments-572773

#### 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 ...

#### 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.

\$2.19