Answer
$18$
Work Step by Step
$\sum ^{3}_{k=0}\left( k^{2}+1\right) =\sum ^{3}_{k=0}k^{2}+\sum ^{3}_{k=0}1=\left( 0^2+\sum ^{3}_{k=1}k^{2}\right) +\left( 1+\sum ^{3}_{k=1}1\right) =\dfrac {n\left( n+1\right) \left( 2n+1\right) }{6}+\left( 1+1\times n\right) =\dfrac {3\times \left( 3+1\right) (2\times 3+1)}{6}+(1+1\times 3)=18$
