# SEQUENCE AND SERIES

A Sequence is a succession of terms arranged in some definite order such that the terms are related to one another according to a well-defined rule.

Examples of sequence
10,  17, 24, 31, 38, ........
-14, -12, -10, -8, .........

The nth of a sequence is represented by $T_n$. Therefore, the first, second, third fourth terms would be represented as $T_1$, $T_2$, $T_3$, $T_4$ respectively.

Some sequences have definite numbers of terms. These are called finite sequences

Some sequences have indefinite numbers of terms. These are called infinite sequences

Example 1
If the nth term of a sequence is $2+4^n$. Find the first three terms of the sequence.

Solution:
$T_n=2+4^n$

When $n=1$
$T_1=2+4^1$
$T_1=2+4$
$T_1=6$

When $n=2$
$T_2=2+4^2$
$T_2=2+16$
$T_2=18$

When $n=3$
$T_3=2+4^3$
$T_3=2+64$
$T_3=63$

Therefore, the first three terms of the sequence are $6$, $18$, and $63$.

Example 2
Given that $T_1=2$, and $T_2=6$, Find the 3rd and 4th term of the sequence if nth terms of the sequence are $6T_{n-1}-2T_{n-2}$

Solution:
$T_n=6T_{n-1}-2T_{n-2}$

$n=3$
$T_3=6T_{3-1}-2T_{3-2}$
$T_3=6T_{2}-2T_{1}$

If $T_1=2$, $T_2=6$, then
$T_3=6(6)-2(2)$
$T_3=36-4$
$T_3=32$

$n=4$
$T_4=6T_{4-1}-2T_{4-2}$
$T_4=6T_{3}-2T_{2}$

By substitution
$T_4=6(32)-2(6)$
$T_4=192-12$
$T_4=180$

Thus, the 3rd and 4th terms of the sequence are $32$ and $180$ respectively.

Example 3
Find the sum of the 3rd and fourth terms of the sequence given by $n(2^n+1)$

Solution:
$Tn=n(2^n+1)$

First of all, we find the value of $T_3$ and $T_4$
$n=3$
$T_3=3(2^3+1)$
$T_3=3(9)$
$T_3=27$

$n=4$
$T_4=4(2^4+1)$
$T_4=4(17)$
$T_4=68$

Therefore, the sum of 3rd and 4ts terms is:
$68+27=95$.

Representation of sequence using factorial notation.

The formula for some sequence contains products or successive positive numbers which can be expressed in factorial notation.

A factorial notation is the product of all the positive numbers preceding or equivalent to $n$.

As an example,

$6!=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$

An example of a sequence formula containing factorial is

$a_n=(n+2)!$

The 3rd term of the sequence above can be found by substituting 3 for n

$a_3=(3+2)!=5!$

$5 × 4 ×3 ×2 ×1=120$

Example 4

Write the first three terms of the sequence $a_n=\frac{3n}{(n+1)!}$

Solution:

Remember that $a_n=\frac{3n}{(n+1)!}$, hence

$a_1=\frac{3(1)}{((1)+1)!}$

$a_1=\frac{3}{2 ×1}$

$a_1=\frac{3}{2}$

$a_2=\frac{3(2)}{((2)+1)!}$

$a_2=\frac{6}{3 ×2 ×1}$

$a_2=\frac{6}{6}=1$

$a_3=\frac{3(3)}{((3)+1)!}$

$a_3=\frac{9}{4 × 3 ×2 ×1}$

$a_3=\frac{9}{24}=\frac{3}{8}$

Hence, the first, second and third terms of the sequence are $\frac{3}{2}$, 1, $\frac{3}{8}$ respectively

Example 5

Write the first three terms of the sequence $a_n=\frac{(n+2)!}{3n}$

Solution:

Remember that $a_n=\frac{(n+2)!}{3n}$

$a_1=\frac{(1+2)!}{3(1)}$

$a_1=\frac{3!}{3}$

$a_1=\frac{3 ×2 ×1}{3}$

$a_1=2$

$a_2=\frac{(2+2)!}{3(2)}$

$a_2=\frac{4!}{6}$

$a_2=\frac{4 × 3 ×2 ×1}{6}$

$a_2=4$

$a_3=\frac{(3+2)!}{3(3)}$

$a_3=\frac{5!}{9}$

$a_3=\frac{5 × 4 × 3 ×2 ×1}{9}$

$a_3=\frac{40}{3}$

The first terms of the sequence are 2, 4, $\frac{40}{3}$

Series

The addition of the terms of a sequence is called a series. The series is usually denoted by $S_n$.

Example 6
Find the series of the first four terms of the sequence whose nth term is $2^{2n}+1$.

Solution:
$T_n=2^{2n}+1$

When $n=1$
$T_1=2^{2(1)}+1$
$T_1=2^{2}+1$
$T_1=5$

When $n=2$
$T_2=2^{2(2)}+1$
$T_2=2^{4}+1$
$T_2=17$

When $n=3$
$T_3=2^{2(3)}+1$
$T_3=2^{6}+1$
$T_3=65$

When $n=4$
$T_4=2^{2(4)}+1$
$T_4=2^{8}+1$
$T_4=257$

Therefore, the required series Is $5+17+65+257......$

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