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