Question 375279
3+5+7...+(2n+1)=n(n+2)
 
n=1 : 2n+1=3, n(n+2)=3
 
 
n->n+1 let 3+5+...+(2n+1) = n(n+2)
 
3+5+...+(2n+1)+(2n+3) = n(n+2) + 2n+3 = n^2 + 2n + 2n + 3 = n^2 +4n+3
 
=(n+3)(n+1)
 
which is n(n+2) with n->n+1, thus the formula is correct.