Power Series and Applications: Background You’ll Need 2

Use the formula for the sum of the first n terms of a geometric series

Geometric Series

The sum of the terms in a geometric sequence is called a geometric series.

formula for the sum of the first [latex]n[/latex] terms of a geometric series

A geometric series is the sum of the terms in a geometric sequence.

The formula for the sum of the first [latex]n[/latex] terms of a geometric sequence is represented as

[latex]{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}[/latex]

How To: Given a geometric series, find the sum of the first [latex]n[/latex] terms.

Identify [latex]{a}_{1},r,\text{ and }n[/latex].
Substitute values for [latex]{a}_{1},r[/latex], and [latex]n[/latex] into the formula [latex]{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}[/latex].
Simplify to find [latex]{S}_{n}[/latex].

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

[latex]{S}_{11}[/latex] for the series [latex]8 + -4 + 2 + \dots[/latex]

[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=\dfrac{-4}{8}=-\frac{1}{2}[/latex]

Substitute values for [latex]{a}_{1}, r, \text{ and } n[/latex] into the formula and simplify.

[latex]\begin{align}&{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r} \\[1mm] &{S}_{11}=\dfrac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336 \\ \text{ } \end{align}[/latex]
[latex]\sum\limits _{k=1}^6 3\cdot {2}^{k}[/latex]

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 [latex]6[/latex], so [latex]n=6[/latex].Substitute values for [latex]{a}_{1},r[/latex], and [latex]n[/latex] into the formula, and simplify.

[latex]\begin{align}\\ &{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r} \\[1mm] &{S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378 \end{align}[/latex]