Apr 7, 2017

$\therefore {\sum}_{k = 1}^{n} \left({k}^{2} - 4 k + 3\right) = \frac{n}{6} \left(n - 1\right) \left(2 n - 7\right) .$

#### Explanation:

We will use $\left(1\right) : {\sum}_{k = 1}^{n} {k}^{2} = \sum {n}^{2} = \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) ,$

$\left(2\right) : \sum n = \frac{n}{2} \left(n + 1\right) , \mathmr{and} , \sum a = a n , \text{ for some const. a.}$

Hence, ${\sum}_{k = 1}^{n} \left({k}^{2} - 4 k + 3\right) ,$

$= {\sum}_{k = 1}^{n} {k}^{2} - {\sum}_{k = 1}^{n} \left(4 k\right) + {\sum}_{k = 1}^{n} 3 ,$

$= \sum {n}^{2} - 4 \sum n + 3 n ,$

$= \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) - 4 \cdot \frac{n}{2} \left(n + 1\right) + 3 n ,$

$= \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) - 2 n \left(n + 1\right) + 3 n ,$

$= \frac{n}{6} \left\{\left(n + 1\right) \left(2 n + 1\right) - 12 \left(n + 1\right) + 18\right\} ,$

$= \frac{n}{6} \left\{2 {n}^{2} + 3 n + 1 - 12 n - 12 + 18\right\} ,$

$= \frac{n}{6} \left(2 {n}^{2} - 9 n + 7\right)$

$\therefore {\sum}_{k = 1}^{n} \left({k}^{2} - 4 k + 3\right) = \frac{n}{6} \left(n - 1\right) \left(2 n - 7\right) .$

Enjoy Maths.!