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+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