Example 1: Using Summation Notation

Evaluate [latex]\sum _{k=3}^{7}{k}^{2}[/latex].

Solution

According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{k}^{2}[/latex] from [latex]k=3[/latex] to [latex]k=7[/latex]. We find the terms of the series by substituting [latex]k=3\text{,}4\text{,}5\text{,}6[/latex], and [latex]7[/latex] into the function [latex]{k}^{2}[/latex]. We add the terms to find the sum.

Example 2: Finding the First n Terms of an Arithmetic Series

Find the sum of each arithmetic series.

1. [latex]\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[/latex]
2. [latex]\text{20 + 15 + 10 +\ldots + -50}[/latex]
3. [latex]\sum _{k=1}^{12}3k - 8[/latex]

Solution

1. We are given [latex]{a}_{1}=5[/latex] and [latex]{a}_{n}=32[/latex].Count the number of terms in the sequence to find [latex]n=10[/latex].Substitute values for [latex]{a}_{1},{a}_{n}\text{\hspace{0.17em},}[/latex] and [latex]n[/latex] into the formula and simplify.
   [latex]\begin{array}{l}\begin{array}{l}\hfill \\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \end{array}\hfill \\ {S}_{10}=\frac{10\left(5+32\right)}{2}=185\hfill \end{array}[/latex]
2. We are given [latex]{a}_{1}=20[/latex] and [latex]{a}_{n}=-50[/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[/latex].
   [latex]\begin{array}{l}{a}_{n}={a}_{1}+\left(n - 1\right)d\hfill \\ -50=20+\left(n - 1\right)\left(-5\right)\hfill \\ -70=\left(n - 1\right)\left(-5\right)\hfill \\ 14=n - 1\hfill \\ 15=n\hfill \end{array}[/latex]
   Substitute values for [latex]{a}_{1},{a}_{n}\text{,}n[/latex] into the formula and simplify.
    
   [latex]\begin{array}{l}\begin{array}{l}\\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\end{array}\hfill \\ {S}_{15}=\frac{15\left(20 - 50\right)}{2}=-225\hfill \end{array}[/latex]
3. To find [latex]{a}_{1}[/latex], substitute [latex]k=1[/latex] into the given explicit formula.
   [latex]\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array}[/latex]
   We are given that [latex]n=12[/latex]. To find [latex]{a}_{12}[/latex], substitute [latex]k=12[/latex] into the given explicit formula.
   [latex]\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array}[/latex]
   Substitute values for [latex]{a}_{1},{a}_{n}[/latex], and [latex]n[/latex] into the formula and simplify.
   [latex]\begin{array}{l}\text{ }{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ {S}_{12}=\frac{12\left(-5+28\right)}{2}=138\hfill \end{array}[/latex]

Example 3: Solving Application Problems with Arithmetic Series

On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?

Solution

This problem can be modeled by an arithmetic series with [latex]{a}_{1}=\frac{1}{2}[/latex] and [latex]d=\frac{1}{4}[/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[/latex], and we are looking for [latex]{S}_{8}[/latex]. To find [latex]{a}_{8}[/latex], we can use the explicit formula for an arithmetic sequence. We can now use the formula for arithmetic series. She will have walked a total of 11 miles.

Example 4: Finding the First n Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.

1. [latex]{S}_{11}[/latex] for the series [latex]\text{ 8 + -4 + 2 + }\dots [/latex]
2. [latex]\underset{6}{\overset{k=1}{{\sum }^{\text{ }}}}3\cdot {2}^{k}[/latex]

Solution

1. [latex]{a}_{1}=8[/latex], and we are given that [latex]n=11[/latex].We can find [latex]r[/latex] by dividing the second term of the series by the first.
   [latex]r=\frac{-4}{8}=-\frac{1}{2}[/latex]
   Substitute values for [latex]{a}_{1}, r, \text{and} n[/latex] into the formula and simplify.
   [latex]\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{11}=\frac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336\hfill \end{array}[/latex]
2. Find [latex]{a}_{1}[/latex] by substituting [latex]k=1[/latex] into the given explicit formula.
   [latex]{a}_{1}=3\cdot {2}^{1}=6[/latex]
   We can see from the given explicit formula that [latex]r=2[/latex]. The upper limit of summation is 6, so [latex]n=6[/latex]. Substitute values for [latex]{a}_{1},r[/latex], and [latex]n[/latex] into the formula, and simplify.
   [latex]\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378\hfill \end{array}[/latex]

Example 5: Solving an Application Problem with a Geometric Series

At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.

Solution

The problem can be represented by a geometric series with [latex]{a}_{1}=26,750[/latex]; [latex]n=5[/latex]; and [latex]r=1.016[/latex]. Substitute values for [latex]{a}_{1}[/latex], [latex]r[/latex], and [latex]n[/latex] into the formula and simplify to find the total amount earned at the end of 5 years. He will have earned a total of $138,099.03 by the end of 5 years.

Example 6: Determining Whether the Sum of an Infinite Series is Defined

Determine whether the sum of each infinite series is defined.

1. [latex]\text{12 + 8 + 4 + }\dots [/latex]
2. [latex]\frac{3}{4}+\frac{1}{2}+\frac{1}{3}+..[/latex].
3. [latex]\sum _{k=1}^{\infty }27\cdot {\left(\frac{1}{3}\right)}^{k}[/latex]
4. [latex]\sum _{k=1}^{\infty }5k[/latex]

Solution

1. The ratio of the second term to the first is [latex]\frac{\text{2}}{\text{3}}[/latex], which is not the same as the ratio of the third term to the second, [latex]\frac{1}{2}[/latex]. The series is not geometric.
2. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of [latex]\frac{2}{3}\text{.}[/latex] The sum of the infinite series is defined.
3. The given formula is exponential with a base of [latex]\frac{1}{3}[/latex]; the series is geometric with a common ratio of [latex]\frac{1}{3}\text{.}[/latex] The sum of the infinite series is defined.
4. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.

Example 7: Finding the Sum of an Infinite Geometric Series

Find the sum, if it exists, for the following:

1. [latex]10+9+8+7+\dots [/latex]
2. [latex]248.6+99.44+39.776+\text{ }\dots [/latex]
3. [latex]\sum _{k=1}^{\infty }4\text{,}374\cdot {\left(-\frac{1}{3}\right)}^{k - 1}[/latex]
4. [latex]\sum _{k=1}^{\infty }\frac{1}{9}\cdot {\left(\frac{4}{3}\right)}^{k}[/latex]

Solution

1. There is not a constant ratio; the series is not geometric.
2. There is a constant ratio; the series is geometric. [latex]{a}_{1}=248.6[/latex] and [latex]r=\frac{99.44}{248.6}=0.4[/latex], so the sum exists. Substitute [latex]{a}_{1}=248.6[/latex] and [latex]r=0.4[/latex] into the formula and simplify to find the sum:
   [latex]\begin{array}{l}S=\frac{{a}_{1}}{1-r}\hfill \\ S=\frac{248.6}{1 - 0.4}=414.\overline{3}\hfill \end{array}[/latex]
3. The formula is exponential, so the series is geometric with [latex]r=-\frac{1}{3}[/latex]. Find [latex]{a}_{1}[/latex] by substituting [latex]k=1[/latex] into the given explicit formula:
   [latex]{a}_{1}=4\text{,}374\cdot {\left(-\frac{1}{3}\right)}^{1 - 1}=4\text{,}374[/latex]
   Substitute [latex]{a}_{1}=4\text{,}374[/latex] and [latex]r=-\frac{1}{3}[/latex] into the formula, and simplify to find the sum:
   [latex]\begin{array}{l}S=\frac{{a}_{1}}{1-r}\hfill \\ S=\frac{4\text{,}374}{1-\left(-\frac{1}{3}\right)}=3\text{,}280.5\hfill \end{array}[/latex]
4. The formula is exponential, so the series is geometric, but [latex]r>1[/latex]. The sum does not exist.

Example 8: Finding an Equivalent Fraction for a Repeating Decimal

Find an equivalent fraction for the repeating decimal [latex]0.\overline{3}[/latex]

Solution

We notice the repeating decimal [latex]0.\overline{3}=0.333..[/latex]. so we can rewrite the repeating decimal as a sum of terms. Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term. Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have

Example 9: Solving an Annuity Problem

A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?

Solution

The value of the initial deposit is $100, so [latex]{a}_{1}=100[/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[/latex]. To find [latex]r[/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit. Substitute [latex]{a}_{1}=100\text{,}r=1.0075\text{,}\text{and}n=120[/latex] into the formula for the sum of the first [latex]n[/latex] terms of a geometric series, and simplify to find the value of the annuity. So the account has $19,351.43 after the last deposit is made.