Answer
$12$
Work Step by Step
According to the provided formula,
$\begin{align}
& N=\frac{1}{2}\left( {{n}^{2}}-n \right) \\
& 66=\frac{1}{2}\left( {{n}^{2}}-n \right) \\
& 132={{n}^{2}}-n
\end{align}$
Subtract $132$ on both sides of the equation,
$\begin{align}
& 132={{n}^{2}}-n \\
& 132-132={{n}^{2}}-n-132 \\
& 0={{n}^{2}}-n-132 \\
& 0={{n}^{2}}-12n+11n-132
\end{align}$
Simplify,
$\begin{align}
& 0=n\left( n-12 \right)+11\left( n-12 \right) \\
& 0=\left( n-12 \right)\left( n+11 \right)
\end{align}$
Thus, the value is $n=12$.
