In this modern era, there is a great increase in the daily life appliances and their buying and selling rate is also increasing day by day. The term sigma notation is widely used in the addition of different prices.

The sum of different appliances can be calculated easily with the help of summation. In this post, we are going to explain the term sigma notation along with its definition, working, kinds, examples, and solutions.

Contents

**Sigma Notation – Definition **

In number theory, the term sigma notation is also known as summation notation. It is a Greek notation that is used to denote the addition of the list of numbers with the initial and final values.

The summation notation is frequently used in hypothesis testing, a sum of squares, variance, and many other similar terms. Now we are going to explain the notation used to denote the summation briefly.

**Notation of summation**

The notation of summation is denoted by the Greek letter “∑”. The subscript of the Greek notation is the initial term and the superscript of the notation is the final value.

j=1m **K**_{j}** = k**_{1}** + k**_{2}** + k**_{3}** + … + k**_{n}

**Two methods of finding the sum**

Here are two methods of finding the sum of the list of numbers.

- Simple summation
- Sigma notation (summation notation)

**Sigma notation**

There is another method of finding a sum other than simple summation is sigma notation. In this method, the function is written with the initial and final values along with the Greek letter sigma.

The sequence of 3y – 2 with the final value of 7 and the initial value is 1.

y=17 [3y – 2] = (3(1) – 2) + (3(2) – 2) + (3(3) – 2) + (3(4) – 2) + (3(5) – 2) + (3(6) – 2) + (3(7) – 2)

y=17 [3y – 2] = (3 – 2) + (6 – 2) + (9 – 2) + (12 – 2) + (15 – 2) + (18 – 2) + (21 – 2)

y=17 [3y – 2] = 1 + 4 + 7 + 10 + 13 + 16 + 19

y=17 [3y – 2] = 70

**Simple summation**

The term simple summation is frequently used to determine the addition of a list of numbers such as the long addition of two or more numbers. For instance, the sum of the first eleven odd numbers can be evaluated easily with the help of simple summation.

Sum of first 11 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Sum of first 11 even numbers = 121

A sigma notation calculator is helpful in finding the result of simple summation and summation notation along with steps.

**Rules of summation notation**

Rule Name | Rule |

Constant Rule | y=1n C = n * C |

Sum Rule | y=1n P_{y} + Q_{y} = y=1n P_{y} + y=1n Q_{y} |

Difference Rule | y=1n P_{y} – Q_{y} = y=1n P_{y} – y=1n Q_{y} |

Product Rule | y=1n P_{y} * Q_{y} = y=1n P_{y} * y=1n Q_{y} |

Quotient Rule | y=1n P_{y} / Q_{y} = y=1n P_{y} / y=1n Q_{y} |

**How to evaluate the sigma notation problems?**

Here is an example of summation to learn how to calculate it manually.

**Example **

Find the sum of the given expression if the starting value is 1 and the ending value is 5.

f(y) = 3y^{4} – 2y^{2} + 2

**Solution **

**Step 1:** Apply the notation of summation to the given function.

y=15 (3y^{4} – 2y^{2} + 2)

**Step 2:** Now apply the given values to the function one by one.

**For y = 1**

3y^{4} – 2y^{2} + 2 = 3(1)^{4} – 2(1)^{2}_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3(1) – 2(1)_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3 – 2 + 2 = 3

**For y = 2**

3y^{4} – 2y^{2} + 2 = 3(2)^{4} – 2(2)^{2}_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3(16) – 2(4)_{ }+ 2

3y^{4} – 2y^{2} + 2 = 48 – 8 + 2 = 42

**For y = 3**

3y^{4} – 2y^{2} + 2 = 3(3)^{4} – 2(3)^{2}_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3(81) – 2(9)_{ }+ 2

3y^{4} – 2y^{2} + 2 = 243 – 18 + 2 = 227

**For y = 4**

3y^{4} – 2y^{2} + 2 = 3(4)^{4} – 2(4)^{2}_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3(256) – 2(16)_{ }+ 2

3y^{4} – 2y^{2} + 2 = 768 – 32 + 2 = 738

**For y = 5**

3y^{4} – 2y^{2} + 2 = 3(5)^{4} – 2(5)^{2}_{ }+ 2

3y^{4} – 2y^{2} + 2 = 3(625) – 2(25)_{ }+ 2

3y^{4} – 2y^{2} + 2 = 1875 – 50 + 2 = 1827

**Step 3:** Calculate the sum of all the above terms.

y=15 (3y^{4} – 2y^{2} + 2) = 3 + 42 + 227 + 738 + 1827

y=15 (3y^{4} – 2y^{2} + 2) = 2837

**Alternately**

**Step 1:** First of all, use the sum and difference law of summation to write them separately.

y=15 (3y^{4} – 2y^{2} + 2) = y=15 (3y^{4}) – y=15 (2y^{2}) + y=15 (2) …. (1)

**Step 2:** Now find the summation of each function.

**For ** y=15 **(3y**^{4}**)**

y=15 (3y^{4}) = 3(1)^{4} + 3(2)^{4} + 3(3)^{4} + 3(4)^{4} + 3(5)^{4}

= 3(1) + 3(16) + 3(81) + 3(256) + 3(625)

= 3 + 48 + 243 + 768 + 1875

= 2937

**For **y=15 **(2y**^{2}**)**

y=15 (2y^{2}) = 2(1)^{2} + 2(2)^{2} + 2(3)^{2} + 2(4)^{2} + 2(5)^{2}

= 2(1) + 2(4) + 2(9) + 2(16) + 2(25)

= 2 + 8 + 18 + 32 + 50

= 110

**For **y=15 **(2)**

y=15 (2) = 2 + 2 + 2 + 2 + 2

= 2 x 5

= 10

**Step 3:** Now put the values in (1).

y=15 (3y^{4} – 2y^{2} + 2) = y=15 (3y^{4}) – y=15 (2y^{2}) + y=15 (2)

= 2937 – 110 + 10

= 2827 + 10

= 2837

**Final Words**

In this article, we have covered all the intent of this post along with a definition, explanation, examples, and solutions. Now you can solve any problem of summation notation just by learning the basics intent of this post.