Chapter 5.2, Problem 36ES

Find the mistakes in the proof fragments in 36-38.

Theorem: For any integer $n\ge 1.$

$1^2+2^2+\mathrm{2...}+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}.$

“Proof (by mathematical induction): Cartainly the theorem is true for $n=1$ because $1^2=1$ and $\frac{1+\left(1+1\right)\left(2\cdot 1+1\right)}{6}=1.$ So the basis
$1^2=1$ and $\frac{1\left(1+1\right)\left(2\cdot +1\right)}{6}=1.$ So the basisstep is true. For the inductive step. Suppose that k is any integer with $k\ge 1,k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}$ . We must show that
${\left(k+1\right)}^2=\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)\left(2\left(k+1\right)+1\right)}{6},$