Problem 99
Question
A deposit of \(P\) dollars is made at the beginning of each month in an account earning an annual interest rate \(r,\) compounded monthly. The balance \(A\) after \(t\) years is given by $$ \begin{aligned} A=P\left(1+\frac{r}{12}\right) &+P\left(1+\frac{r}{12}\right)^{2}+\cdots \\ &+P\left(1+\frac{r}{12}\right)^{12 t} \end{aligned} $$ Show that the balance is given by \(A=P\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right]\left(1+\frac{12}{r}\right)\)
Step-by-Step Solution
Verified Answer
The given balance \(A\) can be represented as seen in a geometric series, and using the sum of a geometric series formula, it can indeed be shown that \(A=P\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right]\left(\frac{12}{r}\right).\).
1Step 1: Recognize the sum as a geometric series
The expression inside the sum is a geometric series with the common ratio being \(\left(1+\frac{r}{12}\right)\). It is important to notice that the series starts from 1, i.e, it is counted from the zeroth power of the common ratio.
2Step 2: Remember the sum formula of a geometric series
The sum (S) of the first \(n\) terms of a geometric series with initial term \(a\) and common ratio \(r\) is given by \(S = a \frac{1-r^n}{1-r}\) if \(r≠1\). In our case, \(a = P\), \(r = \left(1+\frac{r}{12}\right)\) and \(n = 12t\). Now, let's substitute these into the formula.
3Step 3: Substitute into the formula
Substituting \(a = P\), \(r = \left(1+\frac{r}{12}\right)\), and \(n = 12t\) into the geometric series formula, we obtain \[A = P \frac{1- \left(1+\frac{r}{12}\right)^{12t}}{\frac{-r}{12}}\] which simplifies to \[A = P \left(1-\left(1+\frac{r}{12}\right)^{12t}\right)\left(\frac{12}{r}\right)\].
4Step 4: Simplify the expression
By distributing the \(-P \frac{12}{r}\), the formula of the sum becomes \[A = P - P\left(1+\frac{r}{12}\right)^{12t} \left(\frac{12}{r}\right).\] We can then multiply the \(-P \frac{12}{r}\) into the power of \(\left(1+\frac{r}{12}\right)\) and rearrange the formula.
5Step 5: Rearrange the formula
Rearranging the formula yields \[A= P\left[\left(1+\frac{r}{12}\right)^{12t}-1\right]\frac{12}{r},\] which is the same as what we wanted to show. Thus, the original formula for \(A\) can be simplified to this formula.
Key Concepts
Compounded InterestGeometric Series FormulaCompound Interest CalculationAlgebraic Expression Simplification
Compounded Interest
Compounded interest is a method of calculating interest in which interest earned over time is added to the principal amount. This leads to interest being calculated on the new total in future periods, hence compounding the growth of your investment or savings. When interest is compounded monthly, it means the interest is calculated and added to the account balance each month. This results in the account growing faster compared to simple interest, where interest is not compounded.
- Key features of compounded interest include faster accumulation due to the compounding effect.
- The frequency of compounding (daily, monthly, annually) significantly affects the total amount of interest earned.
Geometric Series Formula
A geometric series is a mathematical concept used to sum terms in a sequence where each term is a fixed multiple, known as the common ratio, of the previous term. The formula for the sum of the first \(n\) terms of a geometric series is:
\[S = a \frac{1-r^n}{1-r}\]
where
\[S = a \frac{1-r^n}{1-r}\]
where
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the number of terms.
Compound Interest Calculation
Calculating compound interest involves using the geometric series formula to determine the balance of an account with regular deposits and periodic compounding. Given the initial deposit \(P\), annual interest rate \(r\), and time in years \(t\), when interest is compounded monthly, each month's balance calculation builds on the previous month's balance, shaping a geometric progression.
The balance at the end of \(t\) years can be calculated as
\[A = P \left[ \left(1+\frac{r}{12}\right)^{12t} - 1 \right] \left(\frac{12}{r}\right)\]
This involves
The balance at the end of \(t\) years can be calculated as
\[A = P \left[ \left(1+\frac{r}{12}\right)^{12t} - 1 \right] \left(\frac{12}{r}\right)\]
This involves
- recognizing the monthly compounding factor, \(1+\frac{r}{12}\),
- using the total number of compounding periods, \(12t\).
Algebraic Expression Simplification
Simplifying algebraic expressions involves manipulating and reducing expressions to their most concise and easily interpretable forms. In the context of the exercise, the aim is to break down the cumbersome algebra involved in compounded interest calculations into simpler forms.
By using the geometric series sum formula, you can reduce multiple terms into a more manageable expression. This involves:
By using the geometric series sum formula, you can reduce multiple terms into a more manageable expression. This involves:
- Recognizing patterns within the expression as belonging to a known formula, like the geometric series formula.
- Substituting the appropriate values from the problem into this formula.
- Performing algebraic operations such as distributing terms, factoring, and rearranging terms to simplify the sum to a single expression.
Other exercises in this chapter
Problem 98
Use the Binomial Theorem to expand the complex number. Simplify your result. (Remember that \(i=\sqrt{-1 .})\) \((5-\sqrt{3} i)^{4}\)
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Use the Binomial Theorem to expand the complex number. Simplify your result. (Remember that \(i=\sqrt{-1 .})\) \(\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^
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Use a graphing utility to find the sum. $$\sum_{j=1}^{6}(24-3 j)$$
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