Discrete Mathematics An Open Introduction, 4th Edition

Our goal is to find closed formulas for sequences. Our primary strategy will be to first determine how the sequence is changing from term to term. This will lead to a recurrence relation for the sequence, and from that recurrence relation, we will find a closed formula. We start with two types of sequences that are particularly common and useful: arithmetic and geometric sequences. Along the way, we will explore some techniques for solving recurrence relations.

Suppose you start a business selling prints of mathematical art. In week zero, you sell two prints. Each week after that, you sell four more prints than you did the previous week. How many prints will you sell in the $n$th week?

We call sequences with a constant rate of change arithmetic sequences.

This works for any arithmetic sequence. That is, any sequence with a constant difference will have a linear closed formula, where the “slope” of the linear function is the common difference.

Let’s now argue why the closed formula is correct. One way we could do this is by using a technique sometimes called telescoping (a name which hopefully we become meaningful momentarily).

On the right-hand side, we have added $d$ to itself $n$ times, so the sum is $d\cdot n\text{.}$

The telescoping we did above is useful in other contexts (see Exercise 6), but now that we have established a general form of the closed formula, we can apply it to any arithmetic sequence.

What about sequences like $3,6,12,24,48,\dots \text{?}$ This is not arithmetic because the difference between terms is not constant. However, the ratio between successive terms is constant: $\frac{6}{3}=\frac{12}{6}=\frac{24}{12}=\cdots =2$ We call such sequences geometric.

Recognizing that the sequence is geometric lets us easily write down a recursive definition. $a_n=2a_{n-1}\text{,}$ with $a_0=3\text{.}$

A closed formula is also not difficult to reason out. How do we get the term $a_3$ for example? We start with $3\text{,}$ then multiply by 2 to get $a_1\text{,}$ multiply by $2$ again to get $a_2\text{,}$ and multiply by $2$ a third time to get $a_3\text{.}$ So we multiplied $3$ by $2$ a total of three times, or $a_3=3\cdot 2^3\text{.}$ It looks like $a_n=3\cdot 2^n\text{.}$

Geometric sequences are those which have a growth rate that is proportional to the sequence itself. Just like you might have seen in calculus, it is exactly the exponential functions that have this property.

In the examples and formulas above, we assumed that the initial term was $a_0\text{.}$ If your sequence starts with $a_1\text{,}$ you can easily find the term that would have been $a_0$ and use that in the formula. For example, if we want a formula for the sequence $2,5,8,\dots$ and insist that $2=a_1\text{,}$ then we can find $a_0=-1$ (since the sequence is arithmetic with common difference 3, we have $a_0+3=a_1$). Then the closed formula will be $a_n=-1+3n\text{.}$

Look at the sequence $(T_n{)}_{n\ge 1}$ which starts $1,3,6,10,15,\dots \text{.}$ 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 $3-1=2$ and $6-3=3\ne 2\text{,}$ 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: $2,3,4,5,6,\dots \text{.}$ In other words, the rate of change of this sequence is arithmetic: $T_n-T_{n-1}=n\text{,}$ which immediately gives us the recurrence relation $T_n=T_{n-1}+n\text{.}$

Another way to think of this is that the $n$th term of the sequence $(T_n)$ is the sum of the first $n$ terms in the sequence $1,2,3,4,5,\dots \text{.}$ Thus $(T_n)$ is the sequence of partial sums of the sequence $1,2,3,\dots$ (partial sums because we are not taking the sum of all infinitely many terms).

If we know how to add up the terms of an arithmetic sequence, we can 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 $T_{100}=101\cdot 50=5050\text{.}$

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This insight is usually attributed to Carl Friedrich Gauss, one of the greatest mathematicians of all time, who discovered it as a child when his unpleasant elementary teacher thought he would keep the class busy by requiring them to compute the lengthy sum.

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 $(\genfrac{}{}{0ex}{}{n+1}{2})\text{,}$ 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 $n-1$ 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, which we will explore in the next two sections.