A sequence is a function whose domain is a set of positive integers. For example, the function f(x) = \frac{1}{x^2} \: , x>0 is valid for all values of x greater than 0 which includes decimal numbers. In order to be a sequence, x must take values of positive integers such as 1, 2, 3 and so on. So, sequence is denoted in a slightly different way as

a_n = \frac{1}{n^2} . The first few terms of this sequence can be found as

a_1 = 1, \: a_2 = \frac{1}{4}, \: a_3 = \frac{1}{9} and so on.

Example:

Write first five terms of \left\{c_n\right\} = \left\{(-1)^{n+1}n^2\right\}

\left\{c_1\right\} = 1, \: \left\{c_2\right\} = -4, \: \left\{c_3\right\} = 9 \\ \left\{c_4\right\} = -16, \: \left\{c_5\right\} = 25

A pattern is a series of terms derived from sequence. For example, 1, 3, 5, \ldots \text{is a pattern whose sequence is given by} c_n = 2n - 1

Example:

Write down the nth term from the pattern 1, -2, 3, -4, 5, -6, \ldots

The n th term should be n but with signs. Here even terms are negative and odd terms are positive. So nth terms is

\left\{c_n\right\} = (-1)^{n+1}n

A sequence which is determined by writing first one (or few terms) and then remaining terms are derived from those first few terms are called recursive formula.

Example:

Write the first three terms of the recursive sequence a_1 = 2, \: a_n = 3 + a_{n-1}

a_1 = 2, \\ a_2 = 3 + a_1 = 3 + 2 = 5 \\ a_3 = 3 + a_2 = 3 + 5 = 8

To determine sum of a sequence, summation notation is used to express in compact form.

a_1 + a_2 + a_3 + \ldots + a_n = \sum_{k=1}^{n} a_k

Example:

Express the following sum using summation notation.

\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{13}{13+1}

Note that nth term of the numerator is n while nth term of denominator is n+1. So it can be written as

\sum_{k=1}^{13} \frac{k}{k+1}

Formulas To Calculate Sum of Sequences:

1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \\ \\

1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \\

1^3 + 2^3 + 3^3 + \ldots + n^3 = [\frac{n(n+1)}{2}]^2

Example:

Find the sum of the sequence \sum_{k=1}^{7} (k^2 -3k + 1)

\sum_{k=1}^{7} (k^2 -3k + 1) = \sum_{k=1}^{7} k^2 - 3\sum_{k=1}^{7} k + \sum_{k=1}^{7} 1 \\ \\

= \frac{7(7+1)(2\times7 + 1)}{6} - 3\times \frac{7(7+1)}{2} + 1\times 7 \\

= 140 - 84 + 7 \\

= 63