Calculate $\displaystyle {n}$ if $\displaystyle {\sum_{{{b}={1}}}^{{n}}}{\left({4}{b}-{1}\right)}={666}$ Using the Sigma Laws we get

$\displaystyle {\sum_{{{b}={1}}}^{{n}}}{\left({b}\right)}+$ $\displaystyle ={666}$

But

 $\displaystyle {\sum_{{{b}={1}}}^{{n}}}{\left({b}\right)}=$Preview Question 6 Part 3 of 6   (in terms of $\displaystyle {n}$)

So we have

(in the form $\displaystyle {a}{n}^{{2}}+{b}{n}+{c}={0}$) Preview Question 6 Part 4 of 6  

From which we get

 $\displaystyle {n}=$ (not applicable) or $\displaystyle {n}=$