A. Deriving a Formula for Geometric Series
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A. Deriving a Formula for Geometric Series
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Looking back at the chessboard problem from the Warm Up, write out the first five terms of the series, and find the sum.
\(S_5 = 0.01 + 0.02 + 0.04 + 0.08 + 0.16 = 0.31 \)
When deriving the formula for the sum of an arithmetic series, you added two sets of the same series and then divided by 2, which gave the desired sum.
\[S_n = \frac{{S_n + S_n }}{2} = \frac{{2S_n }}{2} = S_n \]
Nothing was changed because you divided and multiplied by 2, and \(\frac{2}{2} = 1 \). This is one method to manipulate a formula without really changing it. You are going to apply the same strategy as you derive the formula for the sum of a geometric series. However, this time you will multiply and divide by \(\left( {\frac{{r - 1}}{{r - 1}}} \right) = 1\)
\[S_n = S_n \left( {\frac{{r - 1}}{{r - 1}}} \right) = \frac{{rS_n - S_n }}{{r - 1}} = S_n \]
This step may seem pointless here, but observe what happens.
\(S_5 = 0.01 + 0.02 + 0.04 + 0.08 + 0.16 \)
First, multiply the series by the common ratio, \(r = 2 \).
\(r(S_n) = 2S_5 = 0.02 + 0.04 + 0.08 + 0.16 + 0.32 \)
Now, subtract the original series from the ‘doubled’ series. Recall, \(\frac{{rS_n - S_n }}{{r - 1}} = S_n \).
\[\begin{align}
2S_5&&=&&&&&&0.02&&+&&0.04&&+&&0.08&&+&&0.16&&+&&0.32 \\
-(\quad \quad S_5&&=&&0.01&&+&&0.02&&+&&0.04&&+&&0.08&&+&&0.16&&&&&\quad) \\
\hline \\
(2 - 1)S_5 &&= &&-0.01 &&+&& 0 &&+&& 0 &&+&& 0 &&+&& 0 &&+&& 0.32 \\
S_5 &&= &&\frac{0.32 - 0.01}{2 - 1} \\
S_5 &&= &&0.31 \\
\end{align}\]
Using the general term, how does this look?
\[\begin{align}
rS_n &= &&&t_1r &+& t_1r^2 &+& t_1r^3 &+ ... +& t_1 r^{n - 1} &+& t_1r^n \\
-(\quad \qquad S_n &= &t_1 &+& t_1 r &+& t_1 r^2 &+& t_1r^3 &+ ... +& t_1r^{n - 1}&&&&&&) \\
\hline \\
(r - 1)(S_n &= &-t_1 &+& 0 &+& 0 &+& 0 &+ ... +& 0 &+& t_1r^n) \\
S_n &= \frac{t_1r^n - t_1}{r - 1} \\
S_n &= \frac{t_1(r^n - 1)}{r - 1}, {\rm {where}\thinspace} r \ne 1 \\
\end{align}\]
The formula derived above can be used to find the sum of a geometric series. Note that \(r \) cannot equal one, otherwise the denominator would be zero and the sum would be undefined.
\(S_5 = 0.01 + 0.02 + 0.04 + 0.08 + 0.16 = 0.31 \)
When deriving the formula for the sum of an arithmetic series, you added two sets of the same series and then divided by 2, which gave the desired sum.
\[S_n = \frac{{S_n + S_n }}{2} = \frac{{2S_n }}{2} = S_n \]
Nothing was changed because you divided and multiplied by 2, and \(\frac{2}{2} = 1 \). This is one method to manipulate a formula without really changing it. You are going to apply the same strategy as you derive the formula for the sum of a geometric series. However, this time you will multiply and divide by \(\left( {\frac{{r - 1}}{{r - 1}}} \right) = 1\)
\[S_n = S_n \left( {\frac{{r - 1}}{{r - 1}}} \right) = \frac{{rS_n - S_n }}{{r - 1}} = S_n \]
This step may seem pointless here, but observe what happens.
\(S_5 = 0.01 + 0.02 + 0.04 + 0.08 + 0.16 \)
First, multiply the series by the common ratio, \(r = 2 \).
\(r(S_n) = 2S_5 = 0.02 + 0.04 + 0.08 + 0.16 + 0.32 \)
Now, subtract the original series from the ‘doubled’ series. Recall, \(\frac{{rS_n - S_n }}{{r - 1}} = S_n \).
\[\begin{align}
2S_5&&=&&&&&&0.02&&+&&0.04&&+&&0.08&&+&&0.16&&+&&0.32 \\
-(\quad \quad S_5&&=&&0.01&&+&&0.02&&+&&0.04&&+&&0.08&&+&&0.16&&&&&\quad) \\
\hline \\
(2 - 1)S_5 &&= &&-0.01 &&+&& 0 &&+&& 0 &&+&& 0 &&+&& 0 &&+&& 0.32 \\
S_5 &&= &&\frac{0.32 - 0.01}{2 - 1} \\
S_5 &&= &&0.31 \\
\end{align}\]
Using the general term, how does this look?
\[\begin{align}
rS_n &= &&&t_1r &+& t_1r^2 &+& t_1r^3 &+ ... +& t_1 r^{n - 1} &+& t_1r^n \\
-(\quad \qquad S_n &= &t_1 &+& t_1 r &+& t_1 r^2 &+& t_1r^3 &+ ... +& t_1r^{n - 1}&&&&&&) \\
\hline \\
(r - 1)(S_n &= &-t_1 &+& 0 &+& 0 &+& 0 &+ ... +& 0 &+& t_1r^n) \\
S_n &= \frac{t_1r^n - t_1}{r - 1} \\
S_n &= \frac{t_1(r^n - 1)}{r - 1}, {\rm {where}\thinspace} r \ne 1 \\
\end{align}\]
The formula derived above can be used to find the sum of a geometric series. Note that \(r \) cannot equal one, otherwise the denominator would be zero and the sum would be undefined.