Geometric Sequences and Sums
Geometric Sequences
In other words, a sequence is geometric if you can go from one entry to the next simply by repeatedly multiplying the same constant number. Contrast this with the definition of an arithmetic sequence where you are adding the same amount to go from one entry to the next.
For example, the sequence \(M=(1,3,9,27,81,243,…)\) is geometric because to go from \(1\) to \(3\) or from \(3\) to \(9\), or \(9\) to \(27\), or any entry \(a_i\) to the entry right after it, \(a_{i+1}\), all one has to do is multiply the previous entry by the common ratio \(3\) to get the next entry. In this case, one can find this common ratio of \(r=3\) by dividing any element of the sequence by the element right before it (hence the \(a_{i+1}/a_i\) in the box above).
Given this observation, the following describes how we can find the closed formula for a geometric sequence.
Note that in \(A\) we can find each element of \(A\) by multiplying the previous element by the common ratio \(r\). So, if we start from the very first element \(a_0\), we find the second element \(a_1\) by multiplying \(a_0\) by \(r\) (i.e. \(a_1=a_0\cdot r\)). To get from \(a_1\) to \(a_2\), yet again, we multiply \(a_1\) by \(r\) to get \(a_2\) (i.e. \(a_2=a_1\cdot r\)), but this is the same thing as multiplying \(a_0\) by \(r\) twice, no? Similarly, one can find \(a_3\) either by multiplying \(a_2\) by \(r\), or by multiplying \(a_0\) by \(r\) three times. The latter indicates that \(a_3=a_0\cdot r^3\). The numerical example of this just below will help.
Let \(A=(a_0,a_1,a_2,a_3,…)=(2,6,18,54,162,…)\). Notice that the common ratio is \(r=3\). Then notice:
\(2=a_0\)
\(6=a_1=\underline{a_0\cdot r}=2\cdot 3=\underline{a_0\cdot r^1}\)
\(18=a_2=\underline{a_1\cdot r}=6\cdot 3=(2\cdot 3)\cdot 3=2\cdot 3^2=\underline{a_0 \cdot r^2}\)
\(54=a_3=\underline{a_2\cdot r}=18\cdot 3=((2\cdot 3)\cdot 3)\cdot 3=2\cdot 3^3=\underline{a_0\cdot r^3}\)
and so on.
To find the common ratio of a geometric sequence, simply divide any entry by the entry prior. For geometric sequences, this ratio will be the same no matter what pair of entries you pick. Note that \(r=\frac{a_1}{a_0}=\frac{6}{3}=2\) is our common ratio.
The closed formula for \(N\) is thus \(a_i=a_0\cdot r^i=3\cdot 2^i\) because, as in the explanation above, we get from entry to entry by repeatedly multiplying by the common ratio \(r\), which is precisely what is indicated by the exponential expression \(r^i\) or in our case \(2^i\) (since exponentiation is repeated multiplication).
Note that the 23rd element of the sequence is computed by using the closed formula that we found above. In particular,
$$a_{23}=a_0\cdot r^{23}=3\cdot 2^{23}=25165824$$
Finding Sums of (Terminating) Geometric Sequences
Suppose we are working with the (terminating) sequence \(B=(\underline{2},10,50,250,1250,…,3906250,19531250)\). Notice that \(B\) has common ratio \(r=5\) (i.e. you can multiply any entry by \(5\) to get the next entry). With that in mind, what sequence would you get if you multiplied every element of \(B\) by this common ratio?
You would get the sequence (which we will call) \(5B=(10,50,250,1250,6250,…,3906250,19531250,\underline{97656250})\) (again, just multiplied everything in \(B\) by \(5\)).
Notice that everything in \(B\) except \(2\) belongs to \(5B\). Similarly, notice that everything in \(5B\), except \(97656250\), belongs to \(B\). This similarity between these two sets can help us find the sum of all elements of \(B\):
$$S=2+10+50+250+1250+…+3906250+19531250$$
Notice that multiplying \(S\) by the common ratio \(5\) and then subtracting away \(S\) gives us
$$\begin{align}5S&=\ \ \ \ \ \ \ \ \ \ 10+50+250+1250+…+3906250+19531250+97656250\\-S&=-2-10-50-250-1250-…-3906250-19531250\\——&——————————————————————————–\\4S&=-2+\ 0\ +\ 0\ \ +\ 0\ \ +\ \ \ 0\ \ \ +…+\ \ \ \ \ \ \ 0\ \ \ \ \ +\ \ \ \ \ \ \ 0\ \ \ \ \ \ \ +97656250\end{align}$$
In the last line, the \(4S\) came from combining like-terms \(5S\) and \(-S\) on the LHS of the equations. The terms on the RHS of the equation each came from adding straight down the columns.
Now, to finally find our sum \(S\), divide both sides of the equation by that \(4\) to get \(S\) by itself, and then simplify:
$$S=\frac{-2+97656250}{4}=\frac{97656248}{4}=24414062$$
In summary, we were able to find the sum \(S\) of a geometric series by multiplying the sum by the common ratio to get \(5S\), and then cancel out almost every term of the sum \(S\) by subtracting \(S\) from \(5S\). A little division thereafter gives us the value of \(S\).
Common ratio: \(3\)
Closed formula: \(a_i=4\cdot 3^i\)
10th element: \(a_{10}=4\cdot 3^{10}=236196\)
To find the common ratio, simply divide one element by the element that comes before it. Since we are working with a geometric sequence, we will always get the same ration no matter which pair of consecutive entries you pick.
Since the sequence starts at \(4\), and we have to repeatedly multiply by the common ratio \(3\) to go from entry \(0\) to entry \(1\) and from entry \(1\) to entry \(2\) and so on, this is the same as multiplying our starting value of \(4\) by \(3^i\) where \(i\) represents how many sequence entries to the right we wish to jump across; hence the closed formula shown in the solution.
To get the 10th element of the sequence, simply plug \(i=10\) into the closed formula.
Common ratio: \(\frac{1}{2}\)
Closed formula: \(a_i=4\cdot \left(\frac{1}{2}\right)^i\)
12th element: \(a_{10}=4\cdot\left(\frac{1}{2}\right)^{12}=\frac{1}{1024}\)
To find the common ratio, simply divide one element by the element that comes before it. Since we are working with a geometric sequence, we will always get the same ration no matter which pair of consecutive entries you pick.
Since the sequence starts at \(4\), and we have to repeatedly multiply by the common ratio \(\frac{1}{2}\) to go from entry \(0\) to entry \(1\) and from entry \(1\) to entry \(2\) and so on, this is the same as multiplying our starting value of \(4\) by \(\left(\frac{1}{2}\right)^i\) where \(i\) represents how many sequence entries to the right we wish to jump across; hence the closed formula shown in the solution.
To get the 12th element of the sequence, simply plug \(i=12\) into the closed formula.
Sum: \(6291450\)
Sum: \(3985805\)
Sum: \(1.9990234\)ish
Sum: \(2\).