Prove that the sum of consecutive odd numbers beginning at 1 (eg 1, 3, 5, ..) always adds up to a perfect square

Solving by induction

Let P(n) = the proposition at the sum of n the first n odd integer is n^2.

If P(n) was true, then P(n+1) and P(n+k) would have to be true too.

 

1+3+5+...+(2n-1) = n^2

1+3+5+...+(2n-1)+(2n+1) = (n^2) +(2n+1)

                                        = (n^2+2n+1)

                                        = (n+1)^2    True

1+3+5+...+(2n-1)+(2n+1)+(2n+3) = (n^2) +(2n+1)+(2n+3)

                                                   = n^2 + 4n + 3

                                                   = (n+2)^2          True

 

Note: This is an example in my Discrete Math book. I just had a quiz yesterday on induction. :)

Posted by elson on 2004-05-08 06:25:29