1. What is the sequence?

Suppose a real number is supposed with every positive integer n. Then the numbers

are said to form an infinite sequence or simply a sequence, and the numbers themselves are called the terms of the sequence. The following are all examples of sequence.

Example1. The sequence all of whose terms are ones:

Here 1, i.e., the number 1 is associated with every positive integer.

Example 2. The sequence of consecutive odd number:

Here 2n - 1.

Example 3. Consider the sequence specified by the formula

.

Write out the first seven terms of the sequence, and find ,and .

Here is the answer. The first seven terms of the sequence are

.

The nth term of a sequence is called the general term of the sequence. A sequence is often specified by giving an explicit formula for its general term. Thus the sequence such that starts off like

The sequence with general terms is often denoted by {}, i.e., by writing the general term inside curly brackets.

2. Some examples of the sequence.

1)Find the formula for the general term of the sequence

Answer. One possibility is 3n - 1.

Another is (3n - 1) * (the number of digits un n). There are an infinite number of other possibilities!

2) As a less obvious problem, find a formula for the general term of the sequence

Answer. One possibility is .

3. Another example of the sequence.

One could hardly write a resonable formula for the general term of the sequence

Neverthless, the above sequence whose nth term is the nthdigit in the decimal representation of the number Pi= 3.1415926...Hence there is actually a definite rule associating a term of the above sequence with every posotive integer n. For example,=3, =1,etc. Thus, despite the absence of an explicit formula for the genenral term of this sequence, it is possile, at least in principle, to find the number in any given posotion, be it the first, seventh or 1007th position. in particular, it can be shown that the above sequence is not periodic, i.e., that no block of terms repeats itself over and over again like the underlined digits in the decimal expansion of

---

4. Fibonacci Sequence.

Calculate the first 10 digits of the sequence

formed by the following rule: The first terms equal 1(=1, =1), while starting with the third term, every term is the sum of the preceding two terms, i.e.,

.

There is an explicit formula for in this case, but it is not too simple. The terms of the above sequence are called the Fibonacci numbers, and the sequence itself is called the Fibonacci sequence. Here is the spreadsheet of the Fibonacci sequence.

Example. Find the first few terms of the sequence

whose nth term equals the sum of all the positive integers from 1 to n inclusive.

Answer. 1, 3, 6, 10, 15, 21,...

Solution. a1=1, a2=1+2=3, a3=1+2+3=6, a4=a+2+3+4=10,... , =1+2+3+...+n,...

5. Mathematical Induction

If the first person i a line a woman and if there is another woman standing behind every woman (except the last), then every person in the line is a woman. The reasoning behind this somewhat facetious example occurs again and again in mathematics and is called the principle of mathematics induction. We now give a more serious formulation of this principle:

Given a sequence of assertions, if the first assertion is true and if every true assertion is followed by another true assertion, then every assertion in the sequence is true.

Example. Prove that for every positive integer n

1+2+3+ ... + n = . (1)

This formula comprises a whole sequence of assertions:

....

The first assertion of obvious true. We now verify that every true assertion is followed by another true assertion. Suppose assertion k is true, i.e., suppose (1) is valid n=k so that

. (2)

Adding k+1 to both sides of(2), we obtain

But this is just assertion k+1, which comes right after assertion k. Thus we have shown that every true assertion is followed by another true assertion. Hence, according to the principle of mathematical induction, every assertion in the sequence is true, i.e., formula (1) holds for everypositive integer n.

Example The same problem can be solved without recourse to mathematical induction. Writing

1+2+3+ ... +n,

we have

1 + 2 + 3 + ... + (n-2) + (n-1) + n, (3)

and by writing the first sum(3), we get the second sum(4)

n + (n-1) + (n-2) + ... + 3 + 2 + 1, (4).

Adding equations (3) and (4), we find that

2[1+n] + [2+(n-1)] + [3+(n-2)] + ... + [(n-2)+3] + [(n-1)+2] + [n+1].

Each term in brackets equals n+1, and there are exactly n such terms. In other words,

2(n+1) +(n+1) + ... + (n+1) +(n+1) = n(n+1),

and hence

,

which is just another way of writing (1).

6. Another form of the principle of mathematical induction

Another somewhat different form of the principle of mathematical induction goes as follows:

Given any assertion involving an arbitrary positive integer n, suppose that

a) The assertion is true for n=1;

b) Validity of the assertion for n=k implies its validity for n=k+1.

Then assertion is true for every positive integer n.

7. One problem using the principle of mathematical induction.

Prove that is divisible by 5 for every positive integer n.

The proof involves two steps:

a) If n=1, equals 0 and hence is trivially divisible by 5.

b) Let n=k be an arbitary positive integer k, and suppose is divisible by 5. Then is also divisible by 5. In fact, it follows from

that

But each term of the terms on the right is divisible by 5, the first by hypothesis, the second since it is obviously a multiple of 5. Since the sum of two numbers divisible by 5 is itself divisible by 5, it follows thatis divisible by 5. Thus hypothesws a) and b) figuring in the second formulation of the principle of mathematical induction are satisfied. Therefore is divisible by 5 for every positive integer n.

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