Sum of Positive Integers

Introduction about sum of positive integers:

The integers are formed by the natural numbers including 0 (0,1,2,3…..) together with the negatives  of the non-zero natural numbers (-1, -2, -3, ...) The positive integers are 0,1,2,3…..

To find the sum of consecutive positive integers we have to use the following formula,

                 `sum_(i=1)^n`n = `(n (n+1))/(2)`  where n is the positive integer

Properties of integer

Addition (or) sum: If we have added two positive integers the result will be a positive integer. If negative integer involved in addition means the result is based on the biggest number.

Commutative property: a+b= b+a
Associative property:a+(b+c)=(a+b)+c
Identity property: a+1= 1+a

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to solve quadratic inequalities and algebra linear equations. I am sure they will be helpful.

Lets Solve some Examples on Sum of Positive Integers

Pro 1 :Find the sum of consecutive positive integers from 1 to 25

Solution:From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =25  `sum_(i=1)^25`25  = `(25 *(25+1))/(2)`

                                       = `(25 * 26 )/(2)`

                                       =`(650)/(2)`

                                       = 325

The answer for sum of consecutive integer from 1 to 25 is 325

Pro 2:Find the sum of consecutive positive integers from 1 to 50

Solution:From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =50  `sum_(i=1)^50`50  = `(50 *(50+1))/(2)`

                                       = `(50 * 51)/(2)`

                                       =`(2550)/(2)`

                                       = 1275

The answer for sum of consecutive integer from 1 to 25 is 1275

Pro 3:Find the sum of positive integer 5 + 10.

The symbol plus +, represents the positive direction.
So, To find 5+10, number line starts from 5,  go to the  positive direction and move 10 units frontward we will get the 15

5+10=15

The answer is 15

Some more Problems on Sum of Positive Integers

Pro 4: Find the sum of consecutive positive integers from 1 to100

Solution:

From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =100 =  `sum_(i=1)^100`100 = `(100 *(100+1))/(2)`

                                       = `(100 * 101)/(2)`

                                       =`(10100)/(2)`

                                       = 5050

The answer for sum of consecutive integer from 1 to 100 is 5050

Pro 5:Find the sum of consecutive positive integers from 1 to 45

Solution:From the formula `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =45  `sum_(i=1)^45`45  = `(45 *(45+1))/(2)`

                                       =`(45 * 46)/(2)`

                                       =`(2070)/(2)`

                                       = 1035

The answer for sum of consecutive integer from 1 to 45 is 1035