Find the maximum number of spectral lines in Balmer series when an electron returns from the 7th orbit to the 1st orbit in a sample of hydrogen atoms.
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Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$.
Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.
I put together a rough drawing in Inkscape to illustrate all possible transitions*:
I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented:
$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$
For $n = 6$:
$$N = \frac{6(6-1)}{2} = 15$$
Obviously the same result is obtained by taking the sum directly.
* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.
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).