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Geometric Series

Exploring Geometric Series

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

What is the common ratio? $\quad r=$

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

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

$S_n=$

Find the sum of $2+6+18+54+162+486+1458$.

Find the sum of $3+6+12+24+48+\cdots$ to $13$ terms.

Find the sum of $4+12+36+108+\;\cdots\;+8748$.

How many terms are needed for the sum of the geometric series $4 + 12 + 36 + 108 + \cdots$ to exceed $100 000$?

Find the sum of the geometric series that has sixth term $6250$, ninth term $50$, last term $2$.

Consider a geometric series where the sum of the first two terms is $12$ and the sum of the first four terms is $60$. One sequence is $4, 8, 16, 32,\;\ldots$
Write down the first four terms of another sequence.

The sum of the first $n$ terms of a geometric series is given by $S_n = 5^n – 1$.
Find the sixth term of this series.
$u_{6}=$

The sum of the first two terms of a geometric series is $4$ and the sum of the first three terms is $-14$. Find the two possible values for the first term.
Separate your answers with a comma
$u_1=$

Find $n$ given $\displaystyle \sum_{k=1}^n2\times5^{k-1}=39062$
$n=$

A rope of length $6$m is cut into three pieces whose lengths form a geometric sequence.
The longest piece is four times the shortest piece. Find the length of the shortest piece.
Give your answer as a fraction

Josh puts £1000 into an account which earns 10% p.a. interest paid annually.
At the end of each year, Josh puts in another £1000 in the account.
Find the value of the account after 6 years. Round your answer to the nearest £
£

Exercises

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