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Sequences and Series

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In mathematics, the sequence is a collection or list of numbers that have a logical/sequential order or pattern between them. For example, 1, 5, 9, 13, … is a sequence having a difference of 4 between each consecutive next term and each term can be represented in form 1 + 4 * ( n – 1 ) where n is the nth term of the sequence. The sequence can be classified into 3 categories:

  1. Arithmetic Sequence
  2. Geometric Sequence
  3. Harmonic Sequence

Arithmetic Sequence

The sequence in which each consecutive term has a common difference and this difference could be positive, negative and even zero is known as an arithmetic sequence.

Example:

1) 0, 2, 4, 6, …   in this sequence each and every consecutive term has a difference of 2 between them and nth term of sequence can be represented as 2 * ( n – 1 ).

2) 0, 5, 10, 15, …  is another example of arithmetic sequence with a difference of 5 between each consecutive number and nth term of sequence can be represented as 5 * ( n – 1 ).

Geometric Sequence

The sequence in which each consecutive term has a common ratio is known as a Geometric sequence.

Example:

1) 1, 5, 25, 125 … in this sequence, each consecutive term have a ratio of 5 with the term before it and nth term of sequence can be represented as 5 ( n – 1 ).

2) 1, -2, 4, -8, 16, … in this sequence each consecutive term have a ratio of -2 with the term before it and nth term of sequence can be represented as ( -2 )( n – 1 ).

Harmonic Sequence

The sequence in which the reciprocal of each term forms an arithmetic sequence is known as a harmonic sequence.

Example:

(1/5), (1/10), (1/15), (1/20),… in this sequence the reciprocal of each term that is 5, 10, 15, 20, … forms an arithmetic sequence with a difference of 5 between each consecutive term.

What is Summation Notation?

Summation Notation is a simple method to find the sum of a sequence. Summation notation is also known as sigma notation. Sigma refers to the Greek letter sigma, Σ. The limit of the sequence is represented as shown in figure 1 where the lower limit is the starting index of the sequence and the upper limit represents the ending index of the sequence. Like as shown in figure 1 the lower limit is 1 and the upper limit is 4 so this means we need the sum of 1st,2nd,3rd and 4th term which is ( 2 * 1 ) + ( 2 * 2 ) + ( 2 * 3 ) + ( 2 * 4 ) = 2 + 4 + 6 + 8 = 20.

Summation Notation:

\huge \sum\limits_{n=1}^4 (2*n)

Summation Notation for Arithmetic Sequence

Summation Notation of Arithmetic Sequence is of form  Î£ (a + b * n) where a is the first term of the sequence and b is the common difference between any two consecutive terms of the sequence and therefore the nth term of the sequence would be of the form (a + (b * n)).

Example:

Let the arithmetic sequence be 0, 2, 4, 6, … so for making the summation notation we need to find the values of ‘a’ and ‘b’ where ‘a’ is the first term which is 0 so a = 0 and b is the common difference between any 2 consecutive terms which is 2 in this case, so b = 2.Therefore summation notion of sequence would be Σ (0 + (2 * n)) where the lower limit is 0 and the upper limit is ∞ as the first term of the sequence is given as 0 and ending is not defined.

Arithmetic Summation Notation:

\huge \sum\limits_{n=0}^\infin (0\ +\ 2*n)

Summation Notation for Geometric Sequence

Summation Notation of Geometric Sequence is of form  Î£ (a * bn) where a is the first term of the sequence and b is the common ratio between any two consecutive terms of the sequence and therefore the nth term of the sequence would be of the form (a * bn).

Example:

Let the geometric sequence be 2, 10, 50, 250, … so for making the summation notation we need to find the values of ‘a’ and ‘b’ where ‘a’ is the first term which is 2 so a = 2 and b is the common ratio between any 2 consecutive terms which is 5 in this case, so b = 5.Therefore summation notion of sequence would be Σ (2 * 5 n) where the lower limit is 0 and the upper limit is ∞ as the first term of the sequence is given as 0 and ending is not defined.

Geometric Summation Notation:

\huge \sum\limits_{n=0}^\infin (2\ *\ 5^n)

Summation Notation for Harmonic Sequence

Summation Notation of Harmonic Sequence is of form  Î£ ( 1/(a + b*n) )where a is the reciprocal of the first term of the sequence and b is the common difference between reciprocal any two consecutive terms of the sequence and therefore the nth term of the sequence would be of the form (1/(a + b * n)).

Example:

Let the arithmetic sequence be 1/2, 1/4, 1/6, … so for making the summation notation we need to find the values of ‘a’ and ‘b’ where ‘a’ is the reciprocal first term which is 2 so a = 2 and b is the common difference between the reciprocal of any 2 consecutive terms which is 2 in this case, so b = 2.Therefore summation notion of sequence would be Σ (1/( 0 + 2 * n)) where the lower limit is 1 and the upper limit is ∞ as the first term of the sequence is given as 1/2 and ending is not defined. 

Harmonic Summation Notation:

\huge \sum\limits_{n=0}^\infin (\frac{1}{0\ +\ 2^n})

Examples

Example 1: Find the first 4 terms of the sequence: an = 2 * xn + 1 and n > 0?

Solution: 

As we should take care to replace n ( and not x ) with the first 4 natural numbers as n is not equal to 0.

a1 = 2 * x1 + 1 

a2 = 2 * x2 + 1

a3 = 2 * x3 + 1 

a4 = 2 * x4 + 1

Example 2: Find the first 6 terms of the sequence: an \frac{1}{n\ +\ 2} and n ≥ 0?

Solution: 

We have to replace n by the first 6 while numbers (0, 1, 2, 3, 4, 5).

a0 = \frac{1}{0\ +\ 2} = \frac{1}{\ 2}

a1 = \frac{1}{1\ +\ 2} = \frac{1}{\ 3}

a2 = \frac{1}{2\ +\ 2} = \frac{1}{\ 4}

a3 = \frac{1}{3\ +\ 2} = \frac{1}{\ 5}

a4 = \frac{1}{4\ +\ 2} = \frac{1}{\ 6}

a5 = \frac{1}{5\ +\ 2} = \frac{1}{\ 7}

Example 3: Evaluate \sum\limits_{n=1}^k (2*n)  for k = 2 and k = 4?

Solution:

For k = 2

\sum\limits_{n=1}^2 (2*n) = (2 * 1) + (2 * 2) = 2 + 4 = 6 

For k = 4

\sum\limits_{n=1}^4 (2*n)  = (2 * 1) + (2 * 2) + (2 * 3) + (2 * 4) = 2 + 4 + 6 + 8 = 20 

Example 4: Evaluate \sum\limits_{n=1}^4 (5^n)?

Solution:

\sum\limits_{n=1}^4 (5^n) = (51) + (52) + (53) + (54) = 5 + 25 + 125 + 625 = 780

Example 5: Evaluate \sum\limits_{n=0}^3 (\frac{1}{n\ +\ 2^n})?

Solution:

\sum\limits_{n=0}^2 (\frac{1}{n\ +\ 2^n})= (\frac{1}{0\ +\ 2^0})+(\frac{1}{1\ +\ 2^1})+(\frac{1}{2\ +\ 2^2})\\ \sum\limits_{n=0}^2 (\frac{1}{n\ +\ 2^n})=(\frac{1}{1})+(\frac{1}{3})+(\frac{1}{6})\\ \sum\limits_{n=0}^2 (\frac{1}{n\ +\ 2^n})=(\frac{6+2+1}{6})\\ \sum\limits_{n=0}^2 (\frac{1}{n\ +\ 2^n})=(\frac{9}{6})\\ \sum\limits_{n=0}^2 (\frac{1}{n\ +\ 2^n})=(\frac{3}{2})\\

Example 6: Write in expanded form: \sum\limits_{n=0}^\infin (\frac{1}{n\ +\ 2^n}) upto 4 terms?

Solution:

\sum\limits_{n=0}^\infin (\frac{1}{n\ +\ 2^n}) = (\frac{1}{0\ +\ 2^0}) + (\frac{1}{1\ +\ 2^1})+(\frac{1}{2\ +\ 2^2})+(\frac{1}{3\ +\ 2^3})+...\\ \sum\limits_{n=0}^\infin (\frac{1}{n\ +\ 2^n}) = (\frac{1}{1}) + (\frac{1}{3})+(\frac{1}{6})+(\frac{1}{11})+...      



Last Updated : 21 Feb, 2021
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