A shorthand notation for finding sums is to use the Greek capital letter sigma, $\sum$. This is particularly useful in statistics where there are lots of data values to be added together. In our daily life, we come across long sums like $1 + 2 + 3 +….......$ and $1^{2} + 2^{2} + 3^{2} +….......$. It is very difficult to write all the terms in a long sum. So, we represent long sums using a simple form. The Sigma Notation is the easiest way of representing long sums. The symbol used for the sigma notation is "$\sum$" . The Sigma notation helps us to write long sums in a simple way.

1. Usually the sigma notation is represented with numbers below and above the sigma symbols. These numbers show the limits of the values.

2. The Sigma notation has wide application in Statistics. It is used in many formulas in Statistics.

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Sigma, $\sum$ is read as 'sigma', the sigma notation $\sum_{n = 1}^k x_n$ is
referred to as sigma or summation notation and is read as, the sum from $n
= 1$ to $k$. The subscript $n$ is called the index of summation and takes on
the successive values $1, 2, 3, ................, k$. Let $x_n$ be a real number that depends on an integer $n$. Then, the sum of the number is denoted by the symbol $\sum_{n = 1}^k x_n$.

$\sum_{n = 1}^k x_n$ = $x_1 + x_2 + x_3 + ............... + x_k$.

### Sigma Notation:

### Solved Examples

**Question 1: **Solve $sum_{k = 2}^7 K^{2}$

** Solution: **

The expression (using a sigma notation) to be summed here is $k^{2}$.

**Question 2: **

** Solution: **

**Question 3: **Solve the sigma notation of `$\sum_{y=1}^{4}$` (2 + 2y)

** Solution: **

The sigma notation rules will allow us to evaluate formulae containing sigma notation more easily and allow us to derive equivalent formulae.The following are the important Sigma Notation rules:

Let $f(x)$ and $g(x)$ are functions of $x$ and $c$ is a constant, then

1. $\sum c = nc$ , where $n$ is the number of terms in the long sum.

2. $\sum (af(x) + b) = a\sum f(x) + b$

3. $\sum (f + g)(x) = \sum f(x) + \sum g(x)$

1. $1 + 1 + 1 + ..................…. = \sum 1 = n$

2. $1 + 2 + 3 + .....................…. = \sum n =$$ \frac{n(n + 1)}{2}$

3. $1^{2} + 2^{2} + 3^{2} + .....................…. = \sum n^{2} =$ $\frac {n(n + 1)(2n + 1)}{2}$

4. $1^{3} + 2^{3} + 3^{3} + .....................…. = \sum n^{3} = $$(\frac{n(n + 1)}{2})^{2}$

5. Arithmetic Mean = $\frac{\sum x}{n}$

6. Standard Deviation = $\sqrt{\frac{\sum(x - \bar{x})^{2}}{n}}$

There are many formulae in Statistics, which uses Sigma Notation. The above said are only few of them. Sigma is simply an abbreviation to notate long sums. The easiest method to solve the Sigma Notation is by expanding the expression. First, we write all the terms. Then, evaluate each term and add them. This will give the solution of the Sigma notation.

Sigma Notation has wide applications in Arithmetic progression and Geometric Progression.

### Evaluate Sigma Notation

In mathematics, sigma notation is used to find the sum of large number of numbers. It is a way to compactly and precisely express any sum of large numbers. Sigma notation, provides a concise and convenient way of writing long sums.

Lets see with the help of examples how to evaluate the Sigma Notation:

### Solved Examples

**Question 1: **Solve the following Sigma Notation

$\sum_{i=1}^{5}3(i+1)^{2}$

** Solution: **

Given, $\sum_{i=1}^{5}3(i+1)^{2}$

Step 1:

Here each term is of the form 3(i + 1)^{2}.

When we put i = 1, 2, 3…... we get 5 terms of the series. First we evaluate all the terms.

When i = 1, 3(i + 1)^{2} = 3(1 + 1)^{2} = 3(2)^{2} = 12

When i = 2, 3(i + 1)^{2} = 3(2 + 1)^{2} = 3(3)^{2} = 27

When i = 3, 3(i + 1)^{2} = 3(3 + 1)^{2} = 3(4)^{2} = 48

When i = 4, 3(i + 1)^{2} = 3(4 + 1)^{2 }= 3(5)^{2} = 75

When i = 5, 3(i + 1)^{2} = 3(5 + 1)^{2} = 3(6)^{2} = 108

**Step 2:**

=>$\sum_{i = 1}^{5}3(i+1)^{2}$ = 12 + 27 + 48 + 75 + 108 = 270

=> $\sum_{i = 1}^{5}3(i+1)^{2}$ = 270.**answer**

**Answer: 270**

**Question 2: **** **Find the sum of 50 terms in the series 0.2 + 0.22 + 0.222 + …...............

** Solution: **

**Step 1:**

We first take 2 common from all terms. So the long sum becomes

=> 0.2 + 0.22 + 0.222 + …........... = 2[0.1 + 0.11 + 0.111 + …..........]

**Step 2:**

Now we multiply and divide each term by 9

=> $\frac{2}{9}$ * 9[0.1 + 0.11 + 0.111 + …..........] = $\frac{2}{9}$ [0.9 + 0.99 + 0.999 + …]

**Step 3:**

We know that

0.9 = 1 – 0.1,

0.99 = 1 – 0.11,

and so on.

**Step 4:**

=> $\frac{2}{9}$ [0.9 + 0.99 + 0.999 + …] = $\frac{2}{9}$[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + …]

**Step 5:**

Since total number of terms = 50

=> $\frac{2}{9}$[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + …] = $\frac{2}{9} $[50 – (0.1 + 0.01 + 0.001 + …)]

**Step 6:**

{We know that sum to n terms in the geometric progression given by $\frac{a(1-r^{n})}{1-r}$, where 'a' is the first term and 'r' is the common ratio between the terms.

Here a = 0.1 , r = 0.1 and n = 50.

So 0.1 + 0.01 + 0.001 + … = $\frac{0.1(1 - 0.1^{50})}{1 - 0.1}$

So the required sum becomes $\frac{2}{9}[50 -\frac{0.1(1-0.1^{50})}{1-0.1}]$

**Answer:**

Hence the sum of 50 terms in the series 0.2 + 0.22 + 0.222 + …..... is $\frac{2}{9}[50-\frac{0.1(1-0.1^{50})}{1-0.1}]$

The following are some properties of sigma notation.

**Property 1:** $\sum_{n = 1}^k c x_n$ = $c \sum_{n = 1}^k x_n$, where $c$ is any constant.

Property 2: $\sum_{n = 1}^k c$ = $nc$

Property 3:$\sum_{n = 1}^k (x_n + y_n)$ = $\sum_{n = 1}^k x_n$ + $\sum_{n = 1}^k y_n$

Property 4:$\sum_{n = 1}^k (x_n - y_n)$ = $\sum_{n = 1}^k x_n$ - $\sum_{n = 1}^k y_n$

Property 5: $\sum_{n = 1}^k x_n$ = $\sum_{n = 1}^i x_n$ + $\sum_{n = i + 1}^k x_n$, $i < k$, where $i$ and $k$ are positive integers.

Sigma is simply an abbreviation to notate long sums. In double summation, the variable has two subscripts, $X_{ij}$. Here, the first subscript refers to the row that the particular value is in, and the second subscript refers to the column.

$\sum_{j = 1}^m\sum_{i = 1}^n X_{ij} = \sum_{j = 1}^m(x_{1j} + x_{2j} + x_{3j} + ..........+ x_{nj})$

$= (x_{11} + x_{21} + x_{31} + ................. + x_{n1}) + (x_{12} + x_{22} + x_{32} + ................. + x_{n2}) + (x_{13} + x_{23} + x_{33} + ................. + x_{n3}) + ................................... + (x_{1m} + x_{2m} + x_{3m} + ................. + x_{nm})$.

### Solved Example

**Question: **Solve $\sum_{n = 1}^3$$\sum_{m = 1}^2$ $(3^n5^m)$

** Solution: **

Given, $\sum_{n = 1}^3$$\sum_{m = 1}^2$ $(3^n5^m)$

**Step 1:**

$\sum_{n = 1}^3$($\sum_{m = 1}^2$ $(3^n5^m)$)

= $\sum_{n = 1}^3$($3^n5^1 + 3^n5^2)$

**Step 2:**

=**> ** $\sum_{n = 1}^3$($3^n5^1 + 3^n5^2)$ = $(3^1 · 5^1 + 3^1 · 5^2) + (3^2 . 5^1 + 3^2 . 5^2) + (3^3 . 5^1 + 3^3 . 5^2)$

= (15 + 75) + (45 + 225) + (135 + 675)

= 90 + 270 + 810

= 1170

=> $\sum_{n = 1}^3$$\sum_{m = 1}^2$ $(3^n5^m)$ = 1170.**answer**

Lets consider a few examples of Sigma Notation problems:

### Solved Examples

**Question 1: **Write the following sum in the Sigma Notation 1 – 2 + 3 – 4 + 5 – 6 +….........

** Solution: **

Here we can see that there is alternative plus (+) and minus (–) sign. Also with an odd number we have (+) and even number we have (-).

To show the sign of each term we write** **$(-1)^{n-1}$** **in each term.

When n = 1,** **$(-1)^{1-1} = (-1)^{0} = 1$** **

When n = 2, $(-1)^{2 - 1} = (-1)^{1} = -1$

And so on........

So the required sigma notation is** **$\sum (-1)^{n - 1}n$

**Answer:** $\sum (-1)^{n-1}n$

**Question 2: **Write the following sum in the Sigma Notation 1 + 3 + 5 +…..................

** Solution: **

Here we can see that the numbers are odd numbers. So we can write each term as 2n+1.

When n = 0, 2(0) + 1 = 0 + 1 = 1

When n = 1, 2(1) + 1 = 2 + 1 = 3

And so on.

So the required sigma notation is $\sum (2n+1)$.

**Answer: $\sum (2n+1)$**

**Question 3: **** **Expand completely $\sum_{k = 1}^{5}\frac{k}{k+5}$

** Solution: **

Given, $\sum_{k = 1}^{5}\frac{k}{k+5}$

**Step 1:**

Here the limits of long sum are 1 and 5. This means k varies from 1 to 5.

Step 2:

When k=1,

=> $\frac{1}{(1+5)} = \frac{1}{6}$

When k =2,

=> $\frac{2}{(2+5)}$ = $\frac{2}{7}$

and so on.

So the required expansion is $\frac{1}{6}+\frac{2}{7}+\frac{3}{8}+\frac{4}{9}+\frac{5}{10}$

**Answer:** $\frac{1}{6}+\frac{2}{7}+\frac{3}{8}+\frac{4}{9}+\frac{5}{10}$

### Practice Problems

**Question 1: **`$\sum_{k=5}^{500}$` k

Answer : 125240

**Question 2: ** `$\sum_{r=4}^{8}$` (2r+1)

**Question 3: ** `$\sum_{n=1}^{4}$` $\frac{n}{(n+1)}$

$\sum_{n = 1}^k x_n$ = $x_1 + x_2 + x_3 + ............... + x_k$.

Sigma notation is used to represent the sum of numbers in simple form. Suppose the $n^{th}$ term of a series is $u_n$. Then, the sum of the first $20$ terms can be expressed as $\sum_{n = 1}^{20} U_n$. The sum of an arithmetic series is also shown by the sigma notation. For instance, consider the arithmetic series $5 + 8 + 11 + 14 + .......….$ The $n^{th}$ term of this series is $5 + (n - 1)3$, or in a simpler form $3n + 2$. Hence, the sum of the first $20$ terms of the given series can be written as $\sum_{n = 1}^{20} 3n + 2$.

The expression (using a sigma notation) to be summed here is $k^{2}$.

Substitute $k = 2$, then $k = 3......$ up to $k = 7$

Adding the terms together.

$\sum_{k = 2}^7$ k^{2} = 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2}

$= 4 + 9 + 16 + 25 + 36 + 49$

$= 139$

$\sum_{k = 2}^7 K^{2} = 139$

The $n^{th}$ term of an arithmetic sequence is $u_n$ using the sigma notation, where $u_n = 70 - 3n$

(a) Find the values of $u_1$ and $u_2$.

(b) Write the general difference of the arithmetic sequence.

(c) Evaluated $\sum_{n = 1}^30$ u_n$ using the sigma notation.

**Given, **u_{n}= 70 - 3n

(a) Values of u_{1} and u_{2}.

u_{1} = 70 – 3 * 1

= 70 - 3

= 67

=> **u _{1} = 67**

and u_{2}= 70 - 3 * 2

= 70 – 6 = 64

=>** ****u _{2 }= 64**

(b) Common difference (d)

**d =** **u _{2 } - **

= 64 - 67 = - 3**
**

The common difference is therefore -3.

(c) The arithmetic series is 67 + 64 + 61 + … and we need to find the sum of the first 30 terms.

Using the formula S_{n} = $\frac{n}{2}$ [2a +(n-1)d]

With, first term (a) = 67, Common difference (d) = - 3 and number of terms (n) = 30.

S_{30} = $\frac{30}{2}$[134 + 29 x (-3)]

= 15 x 47

= 705

=> Sum of first 30 terms is 705.

Given: `$\sum_{y=1}^{4}$`(2 + 2y)

Substitute the limit to the variable of the given function.

(2 + 2y) = (2 + 2(1)) + (2 + 2(2)) + (2 + 2(3)) + (2 + 2(4))

= (2 + 2) + (2 + 4) + (2 + 6) + (2 + 8)

= 4 + 6 + 8 + 10

= 28

**Answer: ****`sum_(y=1)^4`****(2 + 2y) = 28**

The sigma notation rules will allow us to evaluate formulae containing sigma notation more easily and allow us to derive equivalent formulae.The following are the important Sigma Notation rules:

Let $f(x)$ and $g(x)$ are functions of $x$ and $c$ is a constant, then

1. $\sum c = nc$ , where $n$ is the number of terms in the long sum.

2. $\sum (af(x) + b) = a\sum f(x) + b$

3. $\sum (f + g)(x) = \sum f(x) + \sum g(x)$

To write a long sum in sigma Notation, we
try to find a formula containing a variable $x$. When $x$ is replaced by $1$
in the formula, we get the first term. When we replace $x$ by $2$ in the
formula, we get the second term and so on

1. $1 + 1 + 1 + ..................…. = \sum 1 = n$

2. $1 + 2 + 3 + .....................…. = \sum n =$$ \frac{n(n + 1)}{2}$

3. $1^{2} + 2^{2} + 3^{2} + .....................…. = \sum n^{2} =$ $\frac {n(n + 1)(2n + 1)}{2}$

4. $1^{3} + 2^{3} + 3^{3} + .....................…. = \sum n^{3} = $$(\frac{n(n + 1)}{2})^{2}$

5. Arithmetic Mean = $\frac{\sum x}{n}$

6. Standard Deviation = $\sqrt{\frac{\sum(x - \bar{x})^{2}}{n}}$

There are many formulae in Statistics, which uses Sigma Notation. The above said are only few of them. Sigma is simply an abbreviation to notate long sums. The easiest method to solve the Sigma Notation is by expanding the expression. First, we write all the terms. Then, evaluate each term and add them. This will give the solution of the Sigma notation.

Sigma Notation has wide applications in Arithmetic progression and Geometric Progression.

Lets see with the help of examples how to evaluate the Sigma Notation:

$\sum_{i=1}^{5}3(i+1)^{2}$

Given, $\sum_{i=1}^{5}3(i+1)^{2}$

Step 1:

When we put i = 1, 2, 3…... we get 5 terms of the series. First we evaluate all the terms.

When i = 1, 3(i + 1)

When i = 2, 3(i + 1)

When i = 3, 3(i + 1)

When i = 4, 3(i + 1)

When i = 5, 3(i + 1)

=>

=> $\sum_{i = 1}^{5}3(i+1)^{2}$ = 270.

We first take 2 common from all terms. So the long sum becomes

=> 0.2 + 0.22 + 0.222 + …........... = 2[0.1 + 0.11 + 0.111 + …..........]

Now we multiply and divide each term by 9

=> $\frac{2}{9}$ * 9[0.1 + 0.11 + 0.111 + …..........] = $\frac{2}{9}$ [0.9 + 0.99 + 0.999 + …]

We know that

0.9 = 1 – 0.1,

0.99 = 1 – 0.11,

and so on.

=> $\frac{2}{9}$[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + …] = $\frac{2}{9} $[50 – (0.1 + 0.01 + 0.001 + …)]

{We know that sum to n terms in the geometric progression given by $\frac{a(1-r^{n})}{1-r}$, where 'a' is the first term and 'r' is the common ratio between the terms.

Here a = 0.1 , r = 0.1 and n = 50.

So 0.1 + 0.01 + 0.001 + … = $\frac{0.1(1 - 0.1^{50})}{1 - 0.1}$

So the required sum becomes $\frac{2}{9}[50 -\frac{0.1(1-0.1^{50})}{1-0.1}]$

Hence the sum of 50 terms in the series 0.2 + 0.22 + 0.222 + …..... is $\frac{2}{9}[50-\frac{0.1(1-0.1^{50})}{1-0.1}]$

Property 2:

Property 3:

Property 4:

Property 5:

Sigma is simply an abbreviation to notate long sums. In double summation, the variable has two subscripts, $X_{ij}$. Here, the first subscript refers to the row that the particular value is in, and the second subscript refers to the column.

$\sum_{j = 1}^m\sum_{i = 1}^n X_{ij} = \sum_{j = 1}^m(x_{1j} + x_{2j} + x_{3j} + ..........+ x_{nj})$

$= (x_{11} + x_{21} + x_{31} + ................. + x_{n1}) + (x_{12} + x_{22} + x_{32} + ................. + x_{n2}) + (x_{13} + x_{23} + x_{33} + ................. + x_{n3}) + ................................... + (x_{1m} + x_{2m} + x_{3m} + ................. + x_{nm})$.

Given, $\sum_{n = 1}^3$$\sum_{m = 1}^2$ $(3^n5^m)$

= $\sum_{n = 1}^3$($3^n5^1 + 3^n5^2)$

= (15 + 75) + (45 + 225) + (135 + 675)

= 90 + 270 + 810

= 1170

=> $\sum_{n = 1}^3$$\sum_{m = 1}^2$ $(3^n5^m)$ = 1170.

Here we can see that there is alternative plus (+) and minus (–) sign. Also with an odd number we have (+) and even number we have (-).

To show the sign of each term we write

When n = 1,

So the required sigma notation is

Here we can see that the numbers are odd numbers. So we can write each term as 2n+1.

When n = 0, 2(0) + 1 = 0 + 1 = 1

When n = 1, 2(1) + 1 = 2 + 1 = 3

And so on.

So the required sigma notation is $\sum (2n+1)$.

Given, $\sum_{k = 1}^{5}\frac{k}{k+5}$

Here the limits of long sum are 1 and 5. This means k varies from 1 to 5.

Step 2:

When k=1,

=> $\frac{1}{(1+5)} = \frac{1}{6}$

When k =2,

=> $\frac{2}{(2+5)}$ = $\frac{2}{7}$

and so on.

So the required expansion is $\frac{1}{6}+\frac{2}{7}+\frac{3}{8}+\frac{4}{9}+\frac{5}{10}$

Use appropriate formulae to evaluate the sigma notation.

Answer : 125240

**Answer : **65

**Answer : **$\frac{163}{60}$

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