Introduction to Formula for Sigma:

Sigma is said to be as the function that is adding in series of number. This sigma can be denoted by ∑ using the limit. Example: Let take the function as f(x). The sigma representation of the function is to be added from 1 to n times is denoted as `sum_(x=1)^n`   f(x) = f(1) + f(2) + f(3) --- f(n). In this article, we study about formula for sigma.

 

Formula for Sigma:

Sigma – Formula:

Case 1: Sigma as infinity.

 `sum_(n=1)^m` n =  a = a + a + a + ...... a (m times) = a * (m )

Case 2: Sigma function used as square

  `sum_(i=1)^n` i2 = 12 + 22 + 32 + 42 + 52 + ….. n2 =` (n(n+1)(2n+1))/6`

Case 3: Sigma function used as cube

`sum_(i=1)^n` i3 = 13 + 23 + 33 + 43 + 53 + ….. n3 = `((n^2)(n+1)^2)/4`

These are the formula for the sigma in function. Let us see problem using sigma formula.

Example Problem – Formula for Sigma:

Example 1:

Solve the sigma as  `sum_(x=1)^6`    (x)3

Solution:

Given:  `sum_(x=1)^6`x3

Step 1: Apply the limit from 1 to 6 of given variable of sigma function.

`sum_(c=1)^6`x3 = 13 + 23 + 33 + 43 + 53 + 63

Formula: Summation of cube of infinite number

`sum_(i=1)^n`i3 = 13 + 23 + 33 + 43 + 53 + ….. n3 = `((n^2)(n+1)^2)/4`

= `((6^2)(6+1)^2)/4`

= `(36xx(49))/4`

= `(36 xx 49)/4`

= 9 xx 49 = 441

Answer: x3 = 441

Example 2:

What is the sum of squares of the number using sigma formula 3, 4, 5?

Solution:

Let take the square of number is x

Formula: `sum_(n=1)^5`x2 - `sum_(n=1)^2`x2

`sum_(i=1)^n` `(n(n+1)(2n+1))/6`

Step 1: Find the sum of the squares of the first 5  numbers, that is 1 to 5.

`sum_(n=1)^5`x2 = `(n(n+1)(2n+1))/6`

`sum_(n=1)^5`x2 = `(5xx6xx11)/6`

 `330/6`

= 55

Step 2: Find the sum of squares of first 2numbers using sigma formula

`sum_(n=1)^2`x2 = `(n(n+1)(2n+1))/6`

`sum_(n=1)^2`x2 = `(2xx3xx(4+1))/6`

`sum_(n=1)^2`x2= `(2xx3xx5)/6`

=  `30/6`

`sum_(n=1)^2`= 5

Step 3:  `sum_(n=1)^5`x2 - `sum_(n=1)^2`x2

 55 - 5 = 50

Answer: 50