#### Pens8675309

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Simple Interest vs. Compound Interest (investopedia.com)

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If A is the initial amount invested at 5% per year simple interest, the first year it earns 0.5A interest, the second year it earns 0,5A, etc., That is every year the account earns 0.5A interest so in n years the interest earned will be 0.5nA. In order that the account have doubled the interest earned must be equal to the amount originally in the account: 0.5nA= A. Solve that for n.

With interest compounded annually, the first year it earns 0.5A interest but that interest is add to the account so the interest the second year is 0.5(A+ 0.5A)= 0.5A+ 0.5^2A= A(0.5+0.5^2). The third year the interest earned is 0.5(A+A(0.5+ 0.5^2)= A(0.5+ 0.5+0.5^2+ 0.5^3). The nth year it would be A(0.5+ 0.5^2+ 0.5^3+ ...+ 0.5^n). Including the initial amount in the account, A, that is A(1+0.5+0.5^2+ 0.5^3+ ... + 0.5^n). That sum is a "geometric sum", a sum of the for a+ ar+ ar^2+....+ar^n for some a and r. Here a= 1 and r= 0.5.

Whoever gave you this problem clearly expects you to know how to sum a geometric series. If you don't, or if you need a review, look at Finite Geometric Sum - Bing images .

With interest compounded annually, the first year it earns 0.5A interest but that interest is add to the account so the interest the second year is 0.5(A+ 0.5A)= 0.5A+ 0.5^2A= A(0.5+0.5^2). The third year the interest earned is 0.5(A+A(0.5+ 0.5^2)= A(0.5+ 0.5+0.5^2+ 0.5^3). The nth year it would be A(0.5+ 0.5^2+ 0.5^3+ ...+ 0.5^n). Including the initial amount in the account, A, that is A(1+0.5+0.5^2+ 0.5^3+ ... + 0.5^n). That sum is a "geometric sum", a sum of the for a+ ar+ ar^2+....+ar^n for some a and r. Here a= 1 and r= 0.5.

Whoever gave you this problem clearly expects you to know how to sum a geometric series. If you don't, or if you need a review, look at Finite Geometric Sum - Bing images .

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