Answer
$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-3k+4 \right)}$ is $\text{34}\text{.}$
Work Step by Step
$\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-3k+4 \right)}$
For the sum of the notation,
$\begin{align}
& \sum\limits_{k=0}^{5}{\left( {{k}^{2}}-3k+4 \right)}=\left( {{0}^{2}}-3\cdot 0+4 \right)+\left( {{1}^{2}}-3\cdot 1+4 \right)+\left( {{2}^{2}}-3\cdot 2+4 \right)+\left( {{3}^{2}}-3\cdot 3+4 \right)+\left( {{4}^{2}}-3\cdot 4+4 \right)+\left( {{5}^{2}}-3\cdot 5+4 \right) \\
& =4+2+2+4+8+14 \\
& =34
\end{align}$
Thus, the sum of the sigma notation $\sum\limits_{k=0}^{5}{\left( {{k}^{2}}-3k+4 \right)}$ is$ \ 34 . $
