Sum and Average

Sum of first 'n' number of natural numbers

Choose any set of natural numbers starting with '1'. Multiply the last number to the next number. Then divide the product by two.

Example:

Sum of first 10 natural numbers

= [10(10+1)]/2

= (10*11)/2     

= 110/2           

= 55               

Sum of first 100 natural numbers

= [100(100+1)]/2

= (100*101)/2     

= 10100/2           

= 5050                

Proof

Sn = n/2(a+l)

=> Sn = n/2(1+n)    

=> Sn = n(n+1)/2    

where, Sn = Sum                   

                    n = number of terms

             l = last term = n

Average of first 'n' number of natural numbers

Choose any set of natural numbers starting with '1'. Add 1 to the last term, then divide the result by 2.

Example:

Average of first 10 natural numbers

= (10+1)/2

= 11/2       

= 5.5         

Average of first 100 natural numbers

= (100+1)/2

= 101/2       

= 50.5         

Proof

Average = sum/number of terms

= [n(n+1)/2]/n  

= (n+1)/2         
where, n = number of terms          

Sum of first 'n' number of odd positive integers

Choose any set of odd positive integers starting with '1'. Multiply the number of terms to itself, that is square it.

Example:

Sum of first 10 odd numbers

= (10)^2

= 100    

Sum of first 7 odd numbers

= (7)^2

= 49 

Proof

Sn = n/2[2a+(n-1)d]

=>Sn = n/2[2*1+(n-1)2]   

=>Sn = n/2(2+2n-2)          

=>Sn = n(2n)/2                 

=>Sn = n*n = (n)^2         

where, Sn = sum                              

         n = number of terms

  a = first term = 1

                    d = common difference = 2

Average of first 'n' number of odd positive integers

Average of first 'n' number of odd positive integers is 'n'(the number itself)

Example:

Average of first 10 odd positive integers

= 10

Average of first 7 odd positive integers

= 7

Proof

Average = sum/number of terms

= (n^2)/n

= n          

where, n = number of terms

Sum of first 'n' number of even positive integers

Choose any set of even positive integers starting with '2'. Multiply the number of terms to the next number [n*(n+1)]. The result obtained is the sum.

Example;

Sum of first 10 even numbers

= 10(10+1)

= 10*11 
= 110        

Sum of first 8 even numbers

= 8(8+1)

= 8*9 
= 72   

Proof

Sn = n/2[2a+(n-1)d]

=>Sn = n/2[2*2+(n-1)2]  

=>Sn = n/2(4+2n-2)         

=>Sn = n/2(2n+2)            

=>Sn = n*2(n+1)/2       
=>Sn = n(n+1)                 

where, Sn = sum                              

        n = number of terms

 a = first term = 2

                   d = common difference = 2

Average of first 'n' number of even positive integers

Average of first 'n' number of even positive integers is 'n+1'(the next number)

Example:

Average of first 10 even positive integers

= 10+1
= 11    

Average of first 8 odd positive integers

= 8+1
= 9    

Proof

Average = sum/number of terms

= n(n+1)/n        

= n+1                 

where, n = number of terms         


THANKS FOR READING .................