We take the equation \[1^2+2^2+\cdots+k^2=\frac{1}{6}k(k+1)(2k+1)\nonumber \] (which is \(P(k)\), one of the equations we agreed to suppose was true in the induction hypothesis) and add \((k+1)^2\) to...We take the equation \[1^2+2^2+\cdots+k^2=\frac{1}{6}k(k+1)(2k+1)\nonumber \] (which is \(P(k)\), one of the equations we agreed to suppose was true in the induction hypothesis) and add \((k+1)^2\) to both sides to get \[ \label{indeq1} 1^2+2^2+\cdots+k^2 + (k+1)^2=\frac{1}{6}k(k+1)(2k+1) + (k+1)^2. \] Note that we are trying to prove \(P(k+1)\), so we would like to show that \[1^2+2^2+\cdots+ (k+1)^2=\frac{1}{6}(k+1)(k+1+1)(2(k+1)+1);\nonumber \] we are not quite there yet.
