Question

Find the sum: ${\sum }_{k=4}^{100}(k^2-k+1)$

• A. $4,95,700$
• B. $4,50,100$
• C. $4,20,025$
• D. $4,15,070$

Answer 1

The correct option is A $4,95,700$

First find $a_4$ and $a_{100}$ in equation $(k^2-k+1)$
$a_4=4^2-4+1\Rightarrow 16-4+1\Rightarrow 13$
$a_{100}={100}^2-100+1\Rightarrow 10000-100+1\Rightarrow 9,901$
Then find the sum:
$S_k=\frac{k}{2}(a_4+a_{100})$
$S_{100}=\frac{100}{2}(13+9901)$
$=50(9,914)$
$S_{100}=4,95,700$