What must be added to the sequence 6/2*(1+2)

If a sequence is arithmetic or geometric there are formulas to find the sum of the first n terms, denoted Sn , without actually adding all of the terms.

(Note that a sequence can be neither arithmetic nor geometric, in which case you'll need to add using brute force, or some other strategy.)

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And, yes, it is easier to just add them in this example, as there are only 4 terms. But imagine adding 50 terms ... then the formula is much easier.

Using the Formula

Let's see the formula in action:

Example: Grains of Rice on a Chess Board

On the page Binary Digits we give an example of grains of rice on a chess board. The question is asked:

When we place rice on a chess board:

• 1 grain on the first square,
• 2 grains on the second square,
• 4 grains on the third and so on,
• ...

... doubling the grains of rice on each square ...

... how many grains of rice in total?

So we have:

• a = 1 (the first term)
• r = 2 (doubles each time)
• n = 64 (64 squares on a chess board)

So:

Becomes:

= 1−264−1 = 264 − 1

= 18,446,744,073,709,551,615

Which was exactly the result we got on the Binary Digits page (thank goodness!)

And another example, this time with r less than 1:

Example: Add up the first 10 terms of the Geometric Sequence that halves each time:

{ 1/2, 1/4, 1/8, 1/16, ... }

The values of a, r and n are:

• a = ½ (the first term)
• r = ½ (halves each time)
• n = 10 (10 terms to add)

So:

Becomes:

Very close to 1.

(Question: if we continue to increase n, what happens?)

Why Does the Formula Work?

Let's see why the formula works, because we get to use an interesting "trick" which is worth knowing.

First, call the whole sum "S":  S = a + ar + ar2 + ... + ar(n−2)+ ar(n−1)

Next, multiply S by r:S·r = ar + ar2 + ar3 + ... + ar(n−1) + arn

Notice that S and S·r are similar?

Now subtract them!

Wow! All the terms in the middle neatly cancel out.
(Which is a neat trick)

By subtracting S·r from S we get a simple result:

Let's rearrange it to find S:

Factor out S and a:S(1−r) = a(1−rn)

Divide by (1−r):S = a(1−rn) (1−r)

Which is our formula (ta-da!):

Infinite Geometric Series

So what happens when n goes to infinity?

We can use this formula:

But be careful:

r must be between (but not including) −1 and 1

and r should not be 0 because the sequence {a,0,0,...} is not geometric

So our infnite geometric series has a finite sum when the ratio is less than 1 (and greater than −1)

Let's bring back our previous example, and see what happens:

Example: Add up ALL the terms of the Geometric Sequence that halves each time:

{ 12, 14, 18, 116, ... }

We have:

• a = ½ (the first term)
• r = ½ (halves each time)

And so:

= ½×1½ = 1

Yes, adding 12 + 14 + 18 + ... etc equals exactly 1.

Don't believe me? Just look at this square:

By adding up 12 + 14 + 18 + ...

we end up with the whole thing!

Recurring Decimal

On another page we asked "Does 0.999... equal 1?", well, let us see if we can calculate it:

Example: Calculate 0.999...

We can write a recurring decimal as a sum like this:

And now we can use the formula:

Yes! 0.999... does equal 1.

So there we have it ... Geometric Sequences (and their sums) can do all sorts of amazing and powerful things.

Each number in the sequence is called a term (or sometimes "element" or "member"), read Sequences and Series for a more in-depth discussion.

Finding Missing Numbers

To find a missing number, first find a Rule behind the Sequence.

Sometimes we can just look at the numbers and see a pattern:

Example: 1, 4, 9, 16, ?

Answer: they are Squares (12=1, 22=4, 32=9, 42=16, ...)

Rule: xn = n2

Sequence: 1, 4, 9, 16, 25, 36, 49, ...

Did you see how we wrote that rule using "x" and "n" ?

xn means "term number n", so term 3 is written x3

And we can calculate term 3 using:

x3 = 32 = 9

We can use a Rule to find any term. For example, the 25th term can be found by "plugging in" 25 wherever n is.

x25 = 252 = 625

How about another example:

Example: 3, 5, 8, 13, 21, ?

After 3 and 5 all the rest are the sum of the two numbers before,

That is 3 + 5 = 8, 5 + 8 = 13 etc, which is part of the Fibonacci Sequence:

3, 5, 8, 13, 21, 34, 55, 89, ...

Which has this Rule:

Rule: xn = xn-1 + xn-2

Now what does xn-1 mean? It means "the previous term" as term number n-1 is 1 less than term number n.

And xn-2 means the term before that one.

Let's try that Rule for the 6th term:

x6 = x6-1 + x6-2

x6 = x5 + x4

So term 6 equals term 5 plus term 4. We already know term 5 is 21 and term 4 is 13, so:

x6 = 21 + 13 = 34

Many Rules

One of the troubles with finding "the next number" in a sequence is that mathematics is so powerful we can find more than one Rule that works.

What is the next number in the sequence 1, 2, 4, 7, ?

Here are three solutions (there can be more!):

Solution 1: Add 1, then add 2, 3, 4, ...

So, 1+1=2, 2+2=4, 4+3=7, 7+4=11, etc...

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, ...

(That rule looks a bit complicated, but it works)

Solution 2: After 1 and 2, add the two previous numbers, plus 1:

Rule: xn = xn-1 + xn-2 + 1

Sequence: 1, 2, 4, 7, 12, 20, 33, ...

Solution 3: After 1, 2 and 4, add the three previous numbers

Rule: xn = xn-1 + xn-2 + xn-3

Sequence: 1, 2, 4, 7, 13, 24, 44, ...

So, we have three perfectly reasonable solutions, and they create totally different sequences.

Which is right? They are all right.

... it may be a list of the winners' numbers ... so the next number could be ... anything!

Simplest Rule

When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.

Finding Differences

Sometimes it helps to find the differences between each pair of numbers ... this can often reveal an underlying pattern.

Here is a simple case:

The differences are always 2, so we can guess that "2n" is part of the answer.

Let us try 2n:

n:12345Terms (xn):791113152n:246810Wrong by:55555

The last row shows that we are always wrong by 5, so just add 5 and we are done:

Rule: xn = 2n + 5

OK, we could have worked out "2n+5" by just playing around with the numbers a bit, but we want a systematic way to do it, for when the sequences get more complicated.

Second Differences

In the sequence {1, 2, 4, 7, 11, 16, 22, ...} we need to find the differences ...

... and then find the differences of those (called second differences), like this:

The second differences in this case are 1.

With second differences we multiply by n22

In our case the difference is 1, so let us try just n22:

n:12345Terms (xn):124711n22:0.524.5812.5Wrong by:0.50-0.5-1-1.5

We are close, but seem to be drifting by 0.5, so let us try: n22 − n2

n22 − n2013610Wrong by:11111

Wrong by 1 now, so let us add 1:

n22 − n2 + 1124711Wrong by:00000

We did it!

The formula n22 − n2 + 1 can be simplified to n(n-1)/2 + 1

So by "trial-and-error" we discovered a rule that works:

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, ...

Other Types of Sequences

Read Sequences and Series to learn about:

And there are also:

And many more!

In truth there are too many types of sequences to mention here, but if there is a special one you would like me to add just let me know.

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