**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.

**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` i^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + ….. n^{2}
=` (n(n+1)(2n+1))/6`

**Case 3:** Sigma function used as cube

`sum_(i=1)^n` i^{3} = 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ….. n^{3} = `((n^2)(n+1)^2)/4`

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

Example 1:

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

Solution:

Given: `sum_(x=1)^6`x^{3}

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

`sum_(c=1)^6`x^{3 =} 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3}

Formula: Summation of cube of infinite number

`sum_(i=1)^n`i^{3} = 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + ….. n^{3} = `((n^2)(n+1)^2)/4`

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

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

= `(36 xx 49)/4`

= 9 xx 49 = 441

**Answer:** x^{3} = 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`x^{2} - `sum_(n=1)^2`x^{2}

`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`x^{2} = `(n(n+1)(2n+1))/6`

`sum_(n=1)^5`x^{2} = `(5xx6xx11)/6`

_{ }= `330/6`

= 55

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

`sum_(n=1)^2`x^{2} = `(n(n+1)(2n+1))/6`

`sum_(n=1)^2`x^{2} = `(2xx3xx(4+1))/6`

`sum_(n=1)^2`x^{2}= `(2xx3xx5)/6`

= `30/6`

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

Step 3: `sum_(n=1)^5`x^{2} - `sum_(n=1)^2`x^{2}

55 - 5 = 50

**Answer: 50**