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Mathematics LibreTexts

Supplementary Notes: Sequences, Arithmetic and Geometric

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Investigate!

For the patterns of dots below, draw the next pattern in the sequence. Then give a recursive definition and a closed formula for the number of dots in the nth pattern.

  1. inv_dots-seq1.svg
  2. inv_dots-seq2.svg
  3. inv_dots-seq3.svg

We now turn to the question of finding closed formulas for particular types of sequences.

Arithmetic Sequences

If the terms of a sequence differ by a constant, we say the sequence is arithmetic. If the initial term (a0) of the sequence is a and the common difference is d, then we have,

Recursive definition: an=an1+d with a0=a.

Closed formula: an=a+dn.

How do we know this? For the recursive definition, we need to specify a0. Then we need to express an in terms of an1. If we call the first term a, then a0=a. For the recurrence relation, by the definition of an arithmetic sequence, the difference between successive terms is some constant, say d. So anan1=d, or in other words,

a0=aan=an1+d.

To find a closed formula, first write out the sequence in general:

a0=aa1=a0+d=a+da2=a1+d=a+d+d=a+2da3=a2+d=a+2d+d=a+3d

We see that to find the nth term, we need to start with a and then add d a bunch of times. In fact, add it n times. Thus an=a+dn.

Example SupplementaryNotes.1

Find recursive definitions and closed formulas for the sequences below. Assume the first term listed is a0.

  1. 2,5,8,11,14,.
  2. 50,43,36,29,.
Solution

First we should check that these sequences really are arithmetic by taking differences of successive terms. Doing so will reveal the common difference d.

  1. 52=3, 85=3, etc. To get from each term to the next, we add three, so d=3. The recursive definition is therefore an=an1+3 with a0=2. The closed formula is an=2+3n.
  2. Here the common difference is 7, since we add 7 to 50 to get 43, and so on. Thus we have a recursive definition of an=an17 with a0=50. The closed formula is an=507n.
What about sequences like \(2, 6, 18, 54, \ldots\text{?}\) This is not arithmetic because the difference between terms is not constant. However, the ratio between successive terms is constant. We call such sequences geometric.

The recursive definition for the geometric sequence with initial term a and common ratio r is an=anr;a0=a. To get the next term we multiply the previous term by r. We can find the closed formula like we did for the arithmetic progression. Write

a0=aa1=a0ra2=a1r=a0rr=a0r2

We must multiply the first term a by r a number of times, n times to be precise. We get an=arn.

Geometric Sequences

A sequence is called geometric if the ratio between successive terms is constant. Suppose the initial term a0 is a and the common ratio is r. Then we have,

  • Recursive definition: an=ran1 with a0=a.
  • Closed formula: an=arn.

Example SupplementaryNotes.3

Find the recursive and closed formula for the sequences below. Again, the first term listed is a0.

  1. 3,6,12,24,48,
  2. 27,9,3,1,1/3,
Solution

Again, we should first check that these sequences really are geometric, this time by dividing each term by its previous term. Assuming this ratio is constant, we will have found r.

  1. 6/3=2, 12/6=2, 24/12=2, etc. Yes, to get from any term to the next, we multiply by r=2. So the recursive definition is an=2an1 with a0=3. The closed formula is an=32n.
  2. The common ratio is r=1/3. So the sequence has recursive definition an=13an1 with a0=27 and closed formula an=2713n.

In the examples and formulas above, we assumed that the initial term was a0. If your sequence starts with a1, you can easily find the term that would have been a0 and use that in the formula. For example, if we want a formula for the sequence \(2, 5, 8,\ldots\) and insist that 2=a1, then we can find a0=1 (since the sequence is arithmetic with common difference 3, we have a0+3=a1). Then the closed formula will be an=1+3n.

If you look at other textbooks or online, you might find that their closed formulas for arithmetic and geometric sequences differ from ours. Specifically, you might find the formulas an=a+(n1)d (arithmetic) and an=arn1 (geometric). Which is correct? Both! In our case, we take a to be a0. If instead we had a1 as our initial term, we would get the (slightly more complicated) formulas you find elsewhere.

Sums of Arithmetic and Geometric Sequences

Investigate!

Your neighborhood grocery store has a candy machine full of Skittles.

  1. 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.
    1. How many Skittles will be left in the machine after 20 quarters have been inserted?
    2. Will there ever be exactly zero Skittles left in the machine? Explain.
  2. 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?
  3. 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 (Tn)n1 which starts 1,3,6,10,15,. 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=32, so there is no common difference. Is the sequence geometric? No. 3/1=3 but 6/3=2, so there is no common ratio. What to do?

Notice that the differences between terms form an arithmetic sequence: 2,3,4,5,6,. This says that the nth term of the sequence 1,3,6,10,15, is the sum of the first n terms in the sequence 1,2,3,4,5,. We say that the first sequence is the sequence of partial sums of the second sequence (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 could use this to find a closed formula for a sequence whose differences are the terms of that arithmetic sequence.

This should become clearer if we write the triangular numbers like this:

1=13=1+26=1+2+310=1+2+3+4Tn=1+2+3++n.

Consider how we could find the sum of the first 100 positive integers (that is, T100). Instead of adding them in order, we regroup and add 1+100=101. The next pair to combine is 2+99=101. Then 3+98=101. Keep going. This gives 50 pairs which each add up to 101, so T100=10150=5050. 1 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 Tn=n(n+1)2. Incidentally, this is exactly the same as (n+12), 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.

Summing Arithmetic Sequences: Reverse and Add

Here is a technique that allows us to quickly find the sum of an arithmetic sequence.

Example SupplementaryNotes.4

Find the sum: 2+5+8+11+14++470.

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 second-to-last term also add up to 472. To keep track of everything, we might express this as follows. Call the sum S. Then,

S= 2 + 5 + 8 ++ 467 + 470
+S= 470 + 467 + 464 ++ 5 + 2
2S= 472 + 472 + 472 ++ 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 nth term in the sum (counting 2 as the 0th term) can be expressed as 2+3n. If 2+3n=470 then n=156. 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. Thus

2S=157472=74104

It is now easy to find S:

S=74104/2=37052

This will work for any sum of arithmetic sequences. Call the sum S. Reverse and add. This produces a single number added to itself many times. Find the number of times. Multiply. Divide by 2. Done.

Example SupplementaryNotes.5

Find a closed formula for 6+10+14++(4n2).

Solution

Again, we have a sum of an arithmetic sequence. We need to know how many terms are in the sequence. Clearly each term in the sequence has the form 4k2 (as evidenced by the last term). For which values of k though? To get 6, k=2. To get 4n2 take k=n. So to find the number of terms, we need to know how many integers are in the range 2,3,,n. The answer is n1. (There are n numbers from 1 to n, so one less if we start with 2.)

Now reverse and add:

S= 6 + 10 ++ 4n6 + 4n2
+S= 4n2 + 4n6 ++ 10 + 6
2S= 4n+4 + 4n+4 ++ 4n+4 + 4n+4

Since there are n2 terms, we get

2S=(n2)(4n+4) so S=(n2)(4n+4)2

Besides finding sums, we can use this technique to find closed formulas for sequences we recognize as sequences of partial sums.

Example SupplementaryNotes.6

Use partial sums to find a closed formula for (an)n0 which starts 2,3,7,14,24,37,

Solution

First, if you look at the differences between terms, you get a sequence of differences: 1,4,7,10,13,, which is an arithmetic sequence. Written another way:

a0=2a1=2+1a2=2+1+4a3=2+1+4+7

and so on. We can write the general term of (an) in terms of the arithmetic sequence as follows:

an=2+1+4+7+10++(1+3(n1))

(we use 1+3(n1) instead of 1+3n to get the indices to line up correctly; for a3 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:

an= 2 + 1 + 4 ++ 1+3(n1)
+ an= 2 + 1+3(n1) + 1+3(n2) ++ 1
2an= 4 + 2+3(n1) + 2+3(n1) ++ 2+3(n1)

Not counting the first term (the 4) there are n summands of 2+3(n1)=3n1 so the right-hand side becomes 2+(3n1)n.

Finally, solving for an we get

an=4+(3n1)n2.

Just to be sure, we check a0=42=2, a1=4+22=3, etc. We have the correct closed formula.

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 SupplementaryNotes.7

What is 3+6+12+24++12288?

Solution

Multiply each term by 2, the common ratio. You get 2S=6+12+24++24576. Now subtract: 2SS=3+24576=24573. Since 2SS=S, we have our answer.

To better see what happened in the above example, try writing it this way:

S= 3+ 6+12+24++12288
 2S= 6+12+24++12288 +24576
S= 3+ 0+0+0++0 24576

Then divide both sides by 1 and we have the same result for S. 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 SupplementaryNotes.8

Find a closed formula for S(n)=2+10+50++25n.

Solution

The common ratio is 5. So we have

S =2+10+50++25n
  5S =      10+50++25n+25n+1
4S =225n+1

Thus S=225n+14

Even though this might seem like a new technique, you have probably used it before.

Example SupplementaryNotes.9

Express 0.464646 as a fraction.

Solution

Let N=0.46464646. Consider 0.01N. We get:

N= 0.4646464
0.01N= 0.00464646
0.99N= 0.46

So N=4699. What have we done? We viewed the repeating decimal 0.464646 as a sum of the geometric sequence 0.46,0.0046,0.000046, The common ratio is 0.01. 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.

and notation

To simplify writing out sums, we will use notation like nk=1ak. This means add up the ak's where k changes from 1 to n.

Example SupplementaryNotes.10

Use notation to rewrite the sums:

  1. 1+2+3+4++100
  2. 1+2+4+8++250
  3. 6+10+14++(4n2).
Solution

100k=1k 50k=02k nk=2(4k2)

If we want to multiply the ak instead, we would write nk=1ak. For example, nk=1k=n!.

 


Supplementary Notes: Sequences, Arithmetic and Geometric is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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