1+3+5+...+(2n+1)
adaugam la suma de sus
2+4+6+......+2n
si obtinem
1+2+3+4+5+....+(2n+1)= (2n+1)(2n+2)/2=(2n+1)(n+1) relatia 1
dar
2+4+6+......+2n =2(1+2+3+...+n)=2n(n+1)/2=n(n+1) relatia 2
ca sa obtinem suma initiala scadem din relatia 1 relatia 2
(2n+1)(n+1)-n(n+1)= (n+1)(2n+1-n)=(n+1)(n+1)=(n+1)² deci patrat perfect
