Sum of consecutive odd numbers

 
Parity
http://mathworld.wolfram.com/Parity.html

 

Let     be a positive integer and     be any integer with the same parity as  
with  

These odd integers are all positive.

 

for example,

If   n = 2,    m = 4

8   =   3 + 5

the product is equal to the sum of 2   (n = 2)   consecutive odd integers

 

If   n = 3,   m = 5

15   =   3 + 5 + 7

the product is equal to the sum of 3   (n = 3)   consecutive odd integers

 

If   n = 5,    m = 7

35   =   3 + 5 + 7 + 9 + 11

the product is equal to the sum of 5   (n = 5)   consecutive odd integers

 

Can you prove the product     will always equal to the sum of     consecutive odd integers?

 

                                                             ——————————————
 

Proof   –  

 

The sum of the first     odd integers is   :

From this, we have that the sum of any sequence of consecutive odd integers is :

So an integer     can be written as the sum of consecutive odd integers if it can be expressed as

where

       and       

       or       

       or       

 

  is odd,      is odd :

If  

      

The first odd integer in the sequence is:   2(1) + 1 = 3

The last odd integer in the sequence is   2(4) – 1 = 7

We have the solution  

 
  is even,      is even :

E.g.  

a = 12,    b = 6

n = (a+b)/2         n = (12+6)/2 = 9
m = (a-b)/2         m = (12-6)/2 = 3

1st odd integer is :      2(3+1) – 1 = 7
last odd integer is :      2(9) – 1 = 17

We have the solution   7 + 9 + 11 + 13 + 15 + 17   =   72

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Advertisements