SOLUTION: for any three consecutive numbers prove that the product of the first and third numbers is always one less than the square of the middle number???

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Question 684617: for any three consecutive numbers prove that the product of the first and third numbers is always one less than the square of the middle number???
Answer by Edwin McCravy(20060) About Me  (Show Source):
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for any three consecutive numbers prove that the product of the first and third numbers is always one less than the square of the middle number???
first number = n
middle number = n+1
third number = n+2

product of first and third numbers = n(n+2)
                                   = n²+2n 


one less than the square of the middle number = (n+1)² - 1
                                              = (n+1)(n+1) - 1
                                              = n²+2n+1 - 1
                                              = n²+2n

It's proved because both equal n²+2n

Edwin