$16(x-1)-(x-1)^{2}+(x-1)^{3}-(x-1)^{4}+(x-1)^{5}$ The coefficient of the $x^{2}$ term in the expanded form is $\qquad$
$16-20$. Original expression $=\frac{1}{x}(x-1)\left[1+(x-1)^{5}\right]$, so the coefficient of the $x^{2}$ term in the expansion equals the coefficient of $x^{3}$ in the expansion of $(x-1)^{6}$, which is $C_{6}^{3}(-1)^{3}=-20$.
Another solution: the coefficient of the $x^{2}$ term is $-C_{2}^{0}(-1)^{0}+C_{3}^{1}(-1)-C_{4}^{2}(-1)^{2}+C_{5}^{3} \cdot(-1)^{3}=-20$.