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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?

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bhuvna789456

It is true that the product of two consecutive even integers are always one less than the square of their average.

Step-by-step explanation:

Let the two consecutive odd integers be 1 and 3.

  • The product of 1 and 3 is (1[tex]\times[/tex]3)=3
  • The average of 1 and 3 is (1+3)/2 =4/2 = 2
  • The square of their average is (2)² = 4

∴ The product 3 is one less than the square of their average 4.

Let the two consecutive even integers be 2 and 4.

  • The product of 2 and 4 is (2[tex]\times[/tex]4)=8
  • The average of 2 and 4 is (2+4)/2 =6/2 = 3
  • The square of their average is (3)² = 9

∴ The product 8 is one less than the square of their average 9.

Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.

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