Answer:
Let the to consecutive odd integers be 2n-1 and 2n+1
their average is
=>[tex]\frac{(2n-1) +(2n+1)}{2}[/tex]
=[tex]\frac{4n}{2}[/tex]
= [tex]2n[/tex]
The square of the average is [tex](2n)^2[/tex] = [tex]4n^2[/tex]
One less than the square of their average = [tex]4n^2-1[/tex]-----------------------------(1)
Now the product of the two consecutive odd numbers
(2n-1)( 2n+1) =[tex]4n^2 +2n -2n -1[/tex] = [tex]4n^2 -1[/tex]------------------------------------(2)
From (1) and (2) ,
We can say that the product of two consecutive odd integers is always one less than the square of their average is true
For consecutive even integers
The product of two consecutive integers
(2n)(2n+2) = [tex]4n^2 + 4n[/tex]----------------------------(3)
Whereas, the one less than the square of their average is
= [tex]\frac{(2n) +(2n +2)}{2} - 1[/tex]
= [tex]\frac{(4n +2)}{2} - 1[/tex]
=[tex]2n +1- 1[/tex]
= 2n--------------------------------------------(4)
From (3) and (4) it is clear that product of two consecutive even integers is not one less than the square of their average.
