Recall that (n+1)2=n2+2n+1 and after expansion we get (n+1)2−(2n+1)=n2. Subtract n(2n+1) from both sides (n+1)2−(2n+1)−n(2n+1)=n2−n(2n+1) and rewrite it as (n+1)2−(n+1)(2n+1)=n2−n(2n+1). Now we add (2n+1)24 to both sides: (n+1)2−(n+1)(2n+1)+(2n+1)24=n2−n(2n+1)+(2n+1)24. Factor both sides into square: ((n+1)−2n+12)2=(n−2n+12)2. Now take the square root: (n+1)−2n+12=n−2n+12. Add 2n+12 to both sides and we get n+1=n which is equivalent to 1=0.