To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (foot note 1) spectral lines would be present in the spectrum.
Similarly, when there were more than one atom in the sample, excited electrons ($n_2$) would be in different states $(n_2=2, 3, 4, 5, 6,....,\infty)$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($n_2=8, 7, 6,...$). Suppose those electrons "de-excite" to energy level 2 ($n_1=2$). Thus, electrons in $n_2=8$ can "de-excite" to energy levels $7, 6, 5, 4, 3,$ and $2$ meaning total of 6 spectral lines $(8-2=n_2-n_1)$. Some atoms with electrons in energy level $n_2-1=7$ can also "de-excite" to energy levels $6, 5, 4, 3,$ and $2$ meaning total of 5 spectral lines $(7-2=n_2-1-n_1)$, etc., etc. Thus, total numbers of spectral lines ($s$) in this case would be:
\begin{align}
s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\
&=\frac{(8-2+1)(8-2)}{2}\\
&=\frac{(n_2+1-n_1)(n_2-n_1)}{2}
\end{align}
Foot note 1: Total number of spectral lines for single atom where $n_2=7$ should be: $1 \times 6 = (n_2-n_1)$ in the spectrum, not $n_1(n_2-n_1)$ as I originally suggested (Thanks @porphyrin for careful reading).