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Geometric Series

Exploring Geometric Series

Find the sum of $1+3+9+27+81+243$.

What is the common ratio? $r=$

Multiply each term of the above series by $3$ and write it as a sum.

Let $S_n$ mean the sum of the first $n$ terms of a series. Because our example has 6 terms,
$S_6=1+3+9+27+81+243$

After you multiplied the series by $3$, it became
$3S_6=3+9+27+81+243+729$

Consider subtracting these two equations:
$$
\begin{array}{rc}
3S_6=&&&\color{red}{3}&+&9&+&27&+&81&+&243&+&729\\
S_6=&1&+&\color{red}{3}&+&9&+&27&+&81&+&243\\
\hline
2S_6=
\end{array}
$$

$\color{red}{3}-\color{red}{3}=$

$$
\begin{array}{rc}
3S_6=&&&3&+&\color{red}{9}&+&27&+&81&+&243&+&729\\
S_6=&1&+&3&+&\color{red}{9}&+&27&+&81&+&243\\
\hline
2S_6=&&&0
\end{array}
$$

$\color{red}{9}-\color{red}{9}=$

$$
\begin{array}{rc}
3S_6=&&&3&+&9&+&\color{red}{27}&+&81&+&243&+&729\\
S_6=&1&+&3&+&9&+&\color{red}{27}&+&81&+&243\\
\hline
2S_6=&&&0&+&0
\end{array}
$$

$\color{red}{27}-\color{red}{27}=$

$$
\begin{array}{rc}
3S_6=&&&3&+&9&+&27&+&81&+&243&+&\color{red}{729}\\
S_6=&\color{red}{1}&+&3&+&9&+&27&+&81&+&243\\
\hline
2S_6=&&&0&+&0&+&0&+&0&+&0
\end{array}
$$

$\color{red}{729}-\color{red}{1}=$

$$
\begin{array}{rc}
3S_6=&&&3&+&9&+&27&+&81&+&243&+&729\\
S_6=&1&+&3&+&9&+&27&+&81&+&243\\
\hline
\color{red}{2S_6=}&\color{red}{728}&+&0&+&0&+&0&+&0&+&0
\end{array}
$$

$\color{red}{2S_6=728}\;$ so $\;S_6=$

Sum of a Finite Geometric Series

Let’s find the sum of a general geometric series.
$$
\begin{array}{rc}
S_n=&u_1&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-2}&+&u_1r^{n-1}
\end{array}
$$

If we multiply this series $S_n$ by $r$, we get
$$
\begin{array}{rc}
rS_n=r(u_1&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-2}&+&u_1r^{n-1})
\end{array}
$$

If we multiply $r$ by $u_1$, it becomes

If we multiply $r$ by $u_1r$, it becomes:
*write your answer in the form $u_1r^\square$

If we multiply $r$ by $u_1r^2$, it becomes
*write your answer in the form $u_1r^\square$

If we multiply $r$ by $u_1r^{n-2}$, it becomes
*write your answer in the form $u_1r^\square$

If we multiply $r$ by $u_1r^{n-1}$, it becomes
*write your answer in the form $u_1r^\square$

If we subtract $S_n$ from $rS_n$, we get
$$
\begin{array}{rc}
\color{red}{rS_n}=&&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&+&u_1r^n\\
\color{red}{S_n}=&u_1&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&\\
\hline
\fbox{$\color{red}{rS_n}-\color{red}{S_n}$}=\\
\end{array}
$$

Factorize $S_n$ from $\color{red}{rS_n}-\color{red}{S_n}$.

$$
\begin{array}{rc}
rS_n=&&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&+&\color{red}{u_1r^n}\\
S_n=&\color{red}{u_1}&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&\\
\hline
(r-1)S_n=&\fbox{$\color{red}{u_1r^n}-\color{red}{u_1}$}&+&0&+&0&+&…&+&0\\
\end{array}
$$

Factorize $u_1$ from $\color{red}{u_1r^n}-\color{red}{u_1}$

$$
\begin{array}{rc}
rS_n=&&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&+&u_1r^n\\
S_n=&u_1&+&u_1r&+&u_1r^2&+&…&+&u_1r^{n-1}&\\
\hline
(r-1)S_n=&u_1(r^n-1)\\
\end{array}
$$

$S_n=$
Example 18

Find the sum of $2+6+18+54+…$ to $12$ terms.

Example 18B

Find the sum of $4+12+36+108+…+8748$.

Exercises

(Core 5H on P.120) #1af, 2c, 3ac, 4, 6, 7, 9, 10, 12, 13