Week 7, Day 1

Today we learned about sequences and summation notation.

There are two types of sequences: infinite and finite.

To express an infinite sequence write: a(1), a(2), a(3)... a(n).

These are referred to as the terms of the sequence. 

To express a finite sequence write: a(1), a(2), a(3)... a(n).

If the domain of the function consists of the first "n" positive intergers only, the sequence is finite.

Finding the Terms of a Sequence:

When given a sequence, otherwise known as an explicit formula, such as:

a(n) = 3n - 2 where n = 1, simply plug and chug:

a(1) = 3(1) - 2 = 1

a(2) = 3(2) - 2 = 4

a(3) = 3(3) - 2 = 7

The first three terms of this sequence are 1, 4, and 7.

Finding the nth Term of a Sequence:

When given a set of terms such as:

1, 3, 5, 7, it is most effective to first multiply a number by "n", then add or subtract as necessary:

a(n) = 2n - 1

a(1) = 2(1) - 1 = 1

a(2) = 2(2) - 1 = 3

a(3) = 2(3) - 1 = 5

a(4) = 2(4) - 1 = 7 

A Sequence That Is Defined Reclusively:

The Fibonnaci Sequence is defined reclusively, as follows:

a(0) = 1, a(1) = 1, a(k) = a(k-2) + a(k-1) where k >(=) 2

* i, n, and k are synonymous.

To find the first six terms, simply plug and chug using the sequence provided:

a(0) = 1

a(1) = 1

a(2) = a(0) + a(1) = 1 + 1 = 2

a(3) = a(1) + a(2) = 1 + 2 = 3

a(4) = a(2) + a(3) = 2 + 3 = 5

a(5) = a(3) + a(4) = 3 + 5 = 8

Factorials!:

n! = 1 x 2 x 3 x 4... (n - 1)n 

* 0! = 1

To find 6 factorial, for instance, simply multiply 1, 2, 3, 4, 5, and 6:

6! = 1 x 2 x 3 x 4 x 5 x 6 = 720

To perform factorial operations, for instance 8! / 2!6!, follow the same process:

(1 x 2 x 3 x 4 x 5 x 6 x 7 x 8) / (1 x 2)(1 x 2 x 3 x 4 x 5 x 6) = 28

Summation Notation: