Investigate!
Your neighborhood grocery store has a candy machine full of Skittles.
Suppose that the candy machine currently holds exactly 650 Skittles, and every time someone inserts a quarter, exactly 7 Skittles come out of the machine.
How many Skittles will be left in the machine after 20 quarters have been inserted?
Will there ever be exactly zero Skittles left in the machine? Explain.
What if the candy machine gives 7 Skittles to the first customer who put in a quarter, 10 to the second, 13 to the third, 16 to the fourth, etc. How many Skittles has the machine given out after 20 quarters are put into the machine?
Now, what if the machine gives 4 Skittles to the first customer, 7 to the second, 12 to the third, 19 to the fourth, etc. How many Skittles has the machine given out after 20 quarters are put into the machine?
Look at the sequence \((T_n)_{n\ge 1}\) which starts \(1, 3, 6, 10, 15,\ldots\text{.}\) These are called the triangular numbers since they represent the number of dots in an equilateral triangle (think of how you arrange 10 bowling pins: a row of 4 plus a row of 3 plus a row of 2 and a row of 1).
Is this sequence arithmetic? No, since \(31 = 2\) and \(63 = 3 \ne 2\text{,}\) so there is no common difference. Is the sequence geometric? No. \(3/1 = 3\) but \(6/3 = 2\text{,}\) so there is no common ratio. What to do?
Notice that the differences between terms do form an arithmetic sequence: \(2, 3, 4, 5, 6,\ldots\text{.}\) This means that the \(n\)th term of the sequence \((T_n)\) is the sum of the first \(n\) terms in the sequence \(1,2,3,4,5,\ldots\text{.}\) We say that \((T_n)\) is the sequence of partial sums of the sequence \(1,2,3,\ldots\) (partial sums because we are not taking the sum of all infinitely many terms).
This should become clearer if we write the triangular numbers like this:
\begin{align*}
1 \amp = 1\\
3 \amp = 1+2\\
6 \amp = 1 + 2 + 3\\
10 \amp = 1+2 + 3+ 4\\
\vdots \amp \qquad \vdots\\
T_n \amp = 1 + 2 + 3 + \cdots + n\text{.}
\end{align*}
If we know how to add up the terms of an arithmetic sequence, we could find a closed formula for a sequence whose differences are the terms of that arithmetic sequence. Consider how we could find the sum of the first 100 positive integers (that is, \(T_{100}\)). Instead of adding them in order, we regroup and add \(1+100 = 101\text{.}\) The next pair to combine is \(2+99 = 101\text{.}\) Then \(3+98 = 101\text{.}\) Keep going. This gives 50 pairs which each add up to \(101\text{,}\) so \(T_{100} = 101\cdot 50 = 5050\text{.}\)
In general, using this same sort of regrouping, we find that \(T_n = \frac{n(n+1)}{2}\text{.}\) Incidentally, this is exactly the same as \({n+1 \choose 2}\text{,}\) which makes sense if you think of the triangular numbers as counting the number of handshakes that take place at a party with \(n+1\) people: the first person shakes \(n\) hands, the next shakes an additional \(n1\) hands and so on.
The point of all of this is that some sequences, while not arithmetic or geometric, can be interpreted as the sequence of partial sums of arithmetic and geometric sequences. Luckily there are methods we can use to compute these sums quickly.
Subsubsection Summing Arithmetic Sequences: Reverse and Add
Here is a technique that allows us to quickly find the sum of an arithmetic sequence.
Example 2.2.4.
Find the sum: \(2 + 5 + 8 + 11 + 14 + \cdots + 470\text{.}\)
Solution.
The idea is to mimic how we found the formula for triangular numbers. If we add the first and last terms, we get 472. The second term and secondtolast term also add up to 472. To keep track of everything, we might express this as follows. Call the sum \(S\text{.}\) Then,
\(S =\) 
\(2\) 
\(+\) 
\(5\) 
\(+\) 
\(8\) 
\(+ \cdots +\) 
\(467\) 
\(+\) 
470 
\(+ \quad S =\) 
\(470\) 
\(+\) 
\(467\) 
\(+\) 
\(464\) 
\(+ \cdots +\) 
\(5\) 
\(+\) 
2 
\(2S =\) 
\(472\) 
\(+\) 
\(472\) 
\(+\) 
\(472\) 
\(+ \cdots +\) 
\(472\) 
\(+\) 
\(472\) 
To find \(2S\) then we add 472 to itself a number of times. What number? We need to decide how many terms (summands) are in the sum. Since the terms form an arithmetic sequence, the \(n\)th term in the sum (counting \(2\) as the 0th term) can be expressed as \(2 + 3n\text{.}\) If \(2 + 3n = 470\) then \(n = 156\text{.}\) So \(n\) ranges from 0 to 156, giving 157 terms in the sum. This is the number of 472’s in the sum for \(2S\text{.}\) Thus
\begin{equation*}
2S = 157\cdot 472 = 74104\text{.}
\end{equation*}
It is now easy to find \(S\text{:}\)
\begin{equation*}
S = 74104/2 = 37052\text{.}
\end{equation*}
This will work for the sum of any arithmetic sequence. Call the sum \(S\text{.}\) Reverse and add. This produces a single number added to itself many times. Find the number of times. Multiply. Divide by 2. Done.
Example 2.2.5.
Find a closed formula for \(6 + 10 + 14 + \cdots + (4n  2)\text{.}\)
Solution.
Again, we have a sum of an arithmetic sequence. How many terms are in the sequence? Clearly each term in the sequence has the form \(4k 2\) (as evidenced by the last term). For which values of \(k\) though? To get 6, \(k = 2\text{.}\) To get \(4n2\) take \(k = n\text{.}\) So to find the number of terms, we must count the number of integers in the range \(2,3,\ldots, n\text{.}\) This is \(n1\text{.}\) (There are \(n\) numbers from 1 to \(n\text{,}\) so one less if we start with 2.)
Now reverse and add:
\(S =\) 
\(6\) 
\(+\) 
\(10\) 
\(+ \cdots +\) 
\(4n6\) 
\(+\) 
\(4n2\) 
\(+ \quad S =\) 
\(4n2\) 
\(+\) 
\(4n6\) 
\(+ \cdots +\) 
\(10\) 
\(+\) 
6 
\(2S =\) 
\(4n+4\) 
\(+\) 
\(4n+4\) 
\(+ \cdots +\) 
\(4n+4\) 
\(+\) 
\(4n+4\) 
Since there are \(n1\) terms, we get
\begin{equation*}
2S = (n1)(4n+4)\qquad \mbox{ so } \qquad S = \frac{(n1)(4n+4)}{2}\text{.}
\end{equation*}
Besides finding sums, we can use this technique to find closed formulas for sequences we recognize as sequences of partial sums.
Example 2.2.6.
Use partial sums to find a closed formula for \((a_n)_{n\ge 0}\) which starts \(2, 3, 7, 14, 24, 37,\ldots \ldots\)
Solution.
First, if you look at the differences between terms, you get a sequence of differences: \(1,4,7,10,13, \ldots\text{,}\) which is an arithmetic sequence. Written another way:
\begin{align*}
a_0 \amp = 2\\
a_1 \amp = 2+1\\
a_2 \amp = 2+1+4\\
a_3 \amp = 2+1+4+7
\end{align*}
and so on. We can write the general term of \((a_n)\) in terms of the arithmetic sequence as follows:
\begin{equation*}
a_n = 2 + 1 + 4 + 7 + 10 + \cdots + (1+3(n1))
\end{equation*}
(we use \(1+3(n1)\) instead of \(1+3n\) to get the indices to line up correctly; for \(a_3\) we add up to 7, which is \(1+3(31)\)).
We can reverse and add, but the initial 2 does not fit our pattern. This just means we need to keep the 2 out of the reverse part:
\(a_n =\) 
\(2\) 
\(+\) 
\(1\) 
\(+\) 
\(4\) 
\(+ \cdots +\) 
\(1+3(n1)\) 
\(+ \quad a_n =\) 
\(2\) 
\(+\) 
\(1+3(n1)\) 
\(+\) 
\(1+3(n2)\) 
\(+ \cdots +\) 
\(1\) 
\(2a_n =\) 
\(4\) 
\(+\) 
\(2+3(n1)\) 
\(+\) 
\(2+3(n1)\) 
\(+ \cdots +\) 
\(2+3(n1)\) 
Not counting the first term (the 4) there are \(n\) summands of \(2+3(n1) = 3n1\) so the righthand side becomes \(4+(3n1)n\text{.}\)
Finally, solving for \(a_n\) we get
\begin{equation*}
a_n = \d \frac{4+(3n1)n}{2}\text{.}
\end{equation*}
Just to be sure, we check \(a_0 = \frac{4}{2} = 2\text{,}\) \(a_1 = \frac{4+2}{2} = 3\text{,}\) etc. We have the correct closed formula.
Subsubsection Summing Geometric Sequences: Multiply, Shift and Subtract
To find the sum of a geometric sequence, we cannot just reverse and add. Do you see why? The reason we got the same term added to itself many times is because there was a constant difference. So as we added that difference in one direction, we subtracted the difference going the other way, leaving a constant total. For geometric sums, we have a different technique.
Example 2.2.7.
What is \(3 + 6 + 12 + 24 + \cdots + 12288\text{?}\)
Solution.
Multiply each term by 2, the common ratio. You get \(2S = 6 + 12 + 24 + \cdots + 24576\text{.}\) Now subtract: \(2S  S = 3 + 24576 = 24573\text{.}\) Since \(2S  S = S\text{,}\) we have our answer.
To better see what happened in the above example, try writing it this way:
\(S=\) 
\(3 \, +\) 
\(6 + 12 + 24 + \cdots + 12288\) 

\( \qquad 2S=\) 

\(6 + 12 + 24 + \cdots + 12288\) 
\(+ 24576\) 
\(S = \) 
\(3 \, +\) 
\(0 + 0 + 0 + \cdots + 0 \) 
\(24576\) 
Then divide both sides by \(1\) and we have the same result for \(S\text{.}\) The idea is, by multiplying the sum by the common ratio, each term becomes the next term. We shift over the sum to get the subtraction to mostly cancel out, leaving just the first term and new last term.
Example 2.2.8.
Find a closed formula for \(S(n) = 2 + 10 + 50 + \cdots + 2\cdot 5^n\text{.}\)
Solution.
The common ratio is 5. So we have
\(S\) 
\(= 2 + 10 + 50 + \cdots + 2\cdot 5^n\) 
\( \qquad 5S\) 
\(= ~~~~~~10 + 50 + \cdots + 2\cdot 5^n + 2\cdot5^{n+1}\) 
\(4S\) 
\(= 2  2\cdot5^{n+1}\) 
Thus \(S = \dfrac{22\cdot 5^{n+1}}{4}\)
Even though this might seem like a new technique, you have probably used it before.
Example 2.2.9.
Express \(0.464646\ldots\) as a fraction.
Solution.
Let \(N = 0.46464646\ldots\text{.}\) Consider \(0.01N\text{.}\) We get:
\(N =\) 
\(0.4646464\ldots\) 
\( \qquad 0.01N =\) 
\(0.00464646\ldots\) 
\(0.99N =\) 
\(0.46\) 
So \(N = \frac{46}{99}\text{.}\) What have we done? We viewed the repeating decimal \(0.464646\ldots\) as a sum of the geometric sequence \(0.46, 0.0046, 0.000046, \ldots\) The common ratio is \(0.01\text{.}\) The only real difference is that we are now computing an infinite geometric sum, we do not have the extra “last” term to consider. Really, this is the result of taking a limit as you would in calculus when you compute infinite geometric sums.