Question

Show that the product of two consecutive odd integers is always one less than the square of their average. Is this true also for consecutive even integers?

Answer 1

Answer:

Let the to consecutive odd integers be  2n-1 and 2n+1

their average is

=>$\frac{(2n-1) +(2n+1)}{2}$

=$\frac{4n}{2}$

= $2n$

The square of the average is $(2n)^2$ =  $4n^2$

One less than the square of their average = $4n^2-1$-----------------------------(1)

Now the product of the two consecutive odd numbers

(2n-1)( 2n+1) =$4n^2 +2n -2n -1$ =  $4n^2 -1$------------------------------------(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) =  $4n^2 + 4n$----------------------------(3)

Whereas, the one less than the square of their average is

= $\frac{(2n) +(2n +2)}{2} - 1$

= $\frac{(4n +2)}{2} - 1$

=$2n +1- 1$

= 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.