Answer
$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$ is $\text{43}\text{.}$
Work Step by Step
$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$
For the sum of the notation,
$\begin{align}
& \sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}=\left( {{0}^{2}}-2\cdot 0+3 \right)+\left( {{1}^{2}}-2\cdot 1+3 \right)+\left( {{2}^{2}}-2\cdot 2+3 \right)+\left( {{3}^{2}}-2\cdot 3+3 \right)+\left( {{4}^{2}}-2\cdot 4+3 \right)+\left( {{5}^{2}}-2\cdot 5+3 \right) \\
& =3+2+3+6+11+18 \\
& =43
\end{align}$
Thus, the sum of the sigma notation $\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-2k+3 \right)}$ is $\text{43}\text{.}$
