Recall that \((n+1)^2=n^2+2n+1\). Subtracting \(2n+1\) from both sides gives \((n+1)^2-(2n+1)=n^2\). Now subtract \(n(2n+1)\) from both sides to obtain \((n+1)^2-(2n+1)-n(2n+1)=n^2-n(2n+1)\). Notice that the left-hand side can be rewritten as \((n+1)^2-(n+1)(2n+1)\), so we have \((n+1)^2-(n+1)(2n+1)=n^2-n(2n+1)\).
Next, add \(\frac{(2n+1)^2}{4}\) to both sides. This allows us to complete the square on each side, giving \(((n+1)-\frac{2n+1}{2})^2=(n-\frac{2n+1}{2})^2\).
Taking square roots of both sides leads to \((n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}\). Adding \(\frac{2n+1}{2}\) to both sides produces \(n+1=n\), which simplifies to \(1=0\).