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Revision:OCR Core 2 - Sequences

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 2 - Sequences

1 2. Sequences
2 Also See

2. Sequences

Constructing sequences

A sequence, as one should be aware, is simply a list of numbers which have some form of connection between them, for example:

1, 4, 9, 16, ...

These are the square numbers.

Each value in a sequence is referred to as a term. Sequences are given letters, and one can write a sequence thus:

$u_{0}, \ u_{1}, \ u_{2}, \ ... \ , \ u_{r}$

The choice of the letter "u" for the sequence is merely conventional, and one can expect that other letters will be used. Also, the use of "r" for the "rth term" is arbitrary, however it is commonly used, along with "n".

Sometimes one will begin with the "0th term", and other times with the "1st term", this is why it is useful to refer to a term as $u_{r}$ (where "r" is the number of the term to which one is referring) so as not to obfuscate the meaning in a Mathematical context from that in an "English" context (would "1st term" refer to the "0th term", or the second term, which is $u_{1}$ ?).

One can now think about how one can represent sequences. One does not want to have to write many terms out, however one does not want to leave the reader in any doubt as to the meaning of the sequence, hence one needs to give an expression for the "rth term" (or some other ambiguous term).

Consider the following sequence:

1, 2, 3, 4, 5, 6, ...

One can tell that this sequence is merely one in which one can obtain the next value by adding 1 to the current value. Consider this in terms of sequence notation:

$u_{r + 1} = u_{r} + 1$

(This simply means that for the "(r + 1)th term", one adds 1 to the "rth term").

This is called a recursive definition.

Where does one begin however? This is something that the recursive definition fails to tell us in it's current form. With a recursive definition one needs to have the first term (or some term in the sequence [generally the first]), hence the definition should be revised:

$u_{r + 1} = u_{r} + 1$

$u_{1} = 1$

One should be aware that the method of defining a sequence in this (or some other manner) is necessary to ensure that there is no ambiguity regarding the true meaning of the sequence. One might expect that a sequence will carry on in a certain pattern, but it can, quite possibly, be completely different.

The triangle number sequence

If one is to construct a series of triangles made out of dots, where the first has a single dot, the second has that single dot, and also two other dots; the third has the thee dots from the previous two triangles, and then a further three dots appended, one has a triangular number sequence.

Consider that one can represent this numerically:

$t_{1} = 1$

$t_{2} = 1 + 2$

$t_{3} = 1 + 2 + 3$

Hence:

$t_{r} = 1 + 2 + 3 + \ ... \ + r$

One might now be able to make a recursive definition:

$t_{r + 1} = t_{r} + r + 1$

$t_{1} = 1$

It is more helpful to have a definition which allows one to calculate a given term of the sequence without having to calculate other terms.

Hence one should consider the problem of summing the natural numbers beginning at 1 (as this is what one has seen with the triangular numbers).

To do this, consider the following:

$t = 1 + 2 + 3 + 4 + \ ... \ + r$

$t = r + (r - 1) + (r - 2) + \ ... \ + 1$

(One has merely reversed the order, so the summations are identical [addition is commutative]).

Now consider summing these together:

$2t = (r + 1) + (r + 1) + (r + 1) + \ ... \ + (r + 1)$

There are "r" terms in the sequence, and therefore one can simplify this:

2t = r(r + 1)

Hence:

$t = \frac{r}{2} (r + 1)$ .

This is something that one should be able to recall from memory as it enables one to sum to natural numbers.

Example

1. Sum the natural numbers from 25 to 50 (inclusive), and output the resultant value.

This is a question of the following:

25 + 26 + 27 + ... + 50

Hence one should sum the first fifty natural numbers (beginning at 1), and then subtract the sum of the natural numbers from 1 to 24 (inclusive).

Hence:

r = 50

$\frac{r}{2} (r + 1) = 25 \times 51 = 1275$

Now:

r = 24

$\frac{r}{2} (r + 1) = 12 \times 25 = 300$

Therefore:

1275 - 300 = 975

This is the value that is required.

The factorial sequence

The factorial sequence is linked with the sequence of triangular numbers in the sense that it goes sequentially through the natural numbers, from 1, to a given value; however in the factorial sequence there is multiplication, and not addition, thus giving:

$f_{r + 1) = f_{r} \times (r + 1)$

$f_{0} = 1$

(The reason for defining the term $f_{0}$ as 1 is that one would have a sequence of zeros if it was defined as 0, and also, the applications within the Binomial Theorem have ramifications within the factorial sequence).

Now one can say:

$f_{r} = 1 \times 2 \times 3 \times \ ... \ \times r$

(This may be reversed:

$f_{r} = r \times (r - 1) \times (r - 2) \times \ ... \ \times 1$

There is notation for this sequence (due to it's importance in other areas of Mathematics [notably combinatorics]):

$r! = f_{r} = r \times (r - 1) \times (r - 2) \times \ ... \ \times 1$ .

One should note that 0! = 1.

Example

1. Calculate:

$\displaystyle \frac{1000!}{998!}$ .

Do not multiply these out. It is not possible to calculate either of these values using a pocket calculator, so one must consider the meaning of the fraction:

$\displaystyle \frac{1000!}{998!} = \frac{1000 \times 999 \times 998 \times \ ... \ \times 1}{998 \times 997 \times 996 \times \ ... \ \times 1}$ .

Now one can see that it is possible to cancel the majority of this calculation, leaving:

$\displaystyle \frac{1000!}{998!} = 1000 \times 999 = 999000$ .

Pascal sequences

One will probably have seen Pascal's Triangle, however one might not be aware of some of the patterns and sequences within it.

Consider the following:

1, 1 1, 2, 1 1, 3, 3, 1 1, 4, 6, 4, 1 1, 5, 10, 10, 5, 1

(These are the rows of Pascal's Triangle).

One can use a multiplication based recursive definition to get the terms of the sequence for one row of Pascal's Triangle.

Evidently:

$p_{0} = 1$

Now consider the fifth row:

$p_{1} = 5$

$p_{2} = 10$

$p_{3} = 10$

$p_{4} = 5$

$p_{5} = 1$

One might notice that:

$p_{r + 1} = \frac{5 - r}{r + 1} p_{r}$ .

This will produce a sequence thus:

1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, ...

(One does not put the infinitely many zeros in the triangle, as it would be difficult to represent, and generally these zeros are not considered).

There is a special notation for this:

${n \choose r}$

This is the "rth term" of the "nth row" of Pascal's Triangle. Hence one can see:

${n \choose r + 1} = \frac{n - r}{r + 1} {n \choose r}$ .

Arithmetic sequences

In an arithmetic sequence (otherwise known as an arithmetic progression) the terms of the sequence are separated by constant values.

Consider:

$u = 1, \ 2, \ 3, \ ... \ , \ r$

The concept of a common difference is introduced. In this case, the terms go up in increments of 1, hence the common difference is 1.

This can be generalised:

$u_{r + 1} = u_{r} + d$ .

Now consider how one might calculate a term of the sequence. One does not want to calculate all the terms up to the term which one wishes to know, and therefore one should think about the problem.

If one is to begin with the first term of the sequence:

$u_{1} = a$

$u_{2} = u_{1} + d = a + d$

$u_{3} = u_{2} + d = (a + d) + d = a + 2d$

$u_{r} = u_{r - 1} + d = (a + (r - 2)d) + d = a + (r - 1)d$ .

Example

1. Find the tenth term of the arithmetic progression:

1, 3, 5, ...

One can begin by obtaining the first term, and common difference:

$u_{1} = 1 = a$

5 - 3 = 3 - 1 = 2 = d

r = 10

Therefore:

$u_{10} = 1 + (9 \times 2) = 19$

Finding the sum of arithmetic series

If one has a sequence which is separated by addition symbols, it becomes a series. One often would like to know the total of the terms of an arithmetic sequence, and therefore one should know how to sum the series with which they are associated.

An example of an arithmetic series is:

1 + 2 + 3 + ... + r

There is a common difference, but as this is a series, it is a summation.

One can generalise an arithmetic series:

$a + (a + d) + (a + 2d) + \ ... \ (a + (r - 1)d)$

Now:

$s = a + (a + d) + (a + 2d) + \ ... \ (a + (r - 1)d) = ra + (d + 2d + 3d + 4d + \ ... \ + (r - 1)d)$

Therefore (due to the sum of the natural numbers):

$s = ra + \left[ \frac{r}{2} (r - 1) \right] d$

Now one can factorise:

$s = \frac{r}{2} \left( 2a + (r - 1)d \right)$ .

Example

1. Show that the sum of the natural numbers from 1 to r is:

$\frac{r}{2} (r + 1)$ ,

using the idea of an arithmetic series.

First define the series:

$s = 1 + 2 + 3 + \ ... \ + r$

Hence:

d = 1

a = 1

Therefore:

$s = \frac{r}{2} \left( 2a + (r - 1)d \right) = \frac{r}{2} \left( 2 + (r - 1) \right)$

$s = \frac{r}{2} \left( 2 + r - 1 \right) = \frac{r}{2} (r + 1)$

Sigma notation

It is often convenient to use a special notation for a series (as it is an arduous process to write out several terms, and use an ellipsis).

One can express the following:

$s = u_{1} + u_{2} + u_{3} + \ ... \ + u_{r}$

as:

$\displaystyle \sum_{r = 1}^{r} u_{r}$ .

The "sigma" means "add", the bottom number is the lower limit of the variable, the top number is the upper limit, and the expression succeeding the sigma is representing each term of the series.

This notation is called sigma notation.

One would read the above expression as "sum from 1 to r of $u_{r}$ ".

Example

1. Express the following series using sigma notation:

1 + 2 + 3 + ... + r.

This would be:

$\displaystyle \sum_{r = 1}^{r} r$

This is the sum, from 1 to r, of r.

One should be aware of some rules for sigma notation.

Consider the following:

$\displaystyle \sum_{r = 1}^{r}cu_{r}$

(Where "c" is a real constant).

One can expand this:

$\displaystyle \sum_{r = 1}^{r}cu_{r} = cu_{1} + cu_{2} + cu_{3} + \ ... \ + cu_{r}$

Hence:

$\displaystyle \sum_{r = 1}^{r}cu_{r} = c(u_{1} + u_{2} + u_{3} + \ ... \ + u_{r})$

Therefore:

$\displaystyle \sum_{r = 1}^{r}cu_{r} = c \left[ \sum_{r = 1}^{r}u_{r} \right]$ .

Another rule is to do with addition:

$\displaystyle \sum_{r = 1}^{r} \left( u_{r} + v_{r} \right)$

Now:

$\displaystyle \sum_{r = 1}^{r} \left( u_{r} + v_{r} \right) = u_{1} + v_{1} + u_{2} + v_{2} + u_{3} + v_{3} + \ ... \ + u_{r} + v_{r}$

$\displaystyle \sum_{r = 1}^{r} \left( u_{r} + v_{r} \right) = u_{1} + u_{2} + u_{3} + \ ... \ + u_{r} + v_{1} + v_{2} + v_{3} + \ ... \ + v_{r}$ .

Hence:

$\displaystyle \sum_{r = 1}^{r} \left( u_{r} + v_{r} \right) = \sum_{r = 1}^{r} u_{r} + \sum_{r = 1}^{r} v_{r}$

One should note that the variable "r" in the previous examples is not anything to do with the sequence, it is a dummy variable one could use any letter, but "r" is often used. One might come across a sigma notation that does not use "r = " (or equivalent variable), this has the same meaning.

Sequences with alternating signs

One will come across sequences which have signs that go from positive to negative, or negative to positive.

There is a quick way to deal with the alternating sign, in the form of:

$(-1)^{r}$ ,

or:

$(-1)^{r + 1}$ .

Consider this:

$r = 1, \ 2, \ 3, \ ...$

$(-1)^{r} = -1, \ 1, \ -1, \ ...$

$(-1)^{r + 1} = 1, \ -1, \ 1, \ ...$

This can be applied to a sequence to give the "rth term".

Example

1. Find the "rth term" of the following sequence:

1, -3, 5, -7, ...

One can consider the sequence without the negative signs. This would be a sequence where:

$a = u_{1} = 1$

d = 5 - 3 = 3 - 1 = 2

$u_{r} = 1 + 2(r - 1) = 2r - 1$

Now consider which of the previous expressions should be used for this sequence. This sequence goes from positive to negative, and then back to positive, and so on, so one must use $(-1)^{r + 1}$ .

Hence:

$u_{r} = (-1)^{r + 1} (2r - 1)$ .