The sum of the first $n$ terms in the sequence $8,\:88,\:888,\:8888,\dots$ is ________.

1. $\dfrac{81}{80}\left ( 10^{n}-1 \right )+\dfrac{9}{8}n \\$
2. $\dfrac{81}{80}\left ( 10^{n}-1 \right )-\dfrac{9}{8}n \\$
3. $\dfrac{80}{81}\left ( 10^{n}-1 \right )+\dfrac{8}{9}n \\$
4. $\dfrac{80}{81}\left ( 10^{n}-1 \right )-\dfrac{8}{9}n$

Let $S = 8+88+888+ \ldots +\underbrace{888\dots8}_{n\text { times}}$

$\implies S = 8\left [1+11+111+ \ldots + \underbrace{111\dots 1}_{n\text { times}} \right]$

$\implies S = \dfrac{8}{9}\left [9+99+999+ \ldots + \underbrace{999\dots 9}_{n\text { times}}\right]$

$\implies S = \dfrac{8}{9}\left [(10-1)+(10^{2}-1)+(10^{3}-1)+ \ldots + (10^{n} – 1) \right]$

$\implies S = \dfrac{8}{9}\left [\underbrace{10^{1} + 10^{2} + 10^{3}+ \ldots + 10^{n}}_{\textbf{GP Series}}\:- n \right]$

$\implies S = \dfrac{8}{9}\left[\dfrac{10(10^{n} - 1)}{10-1} - n \right]\:\:\:\:\left[\because\text{Sum of GP}: S_{n} = \dfrac{a(r^{n} - 1)}{r-1}\:\:; r>0\:\: \text{and}\:\: S_{n} = \dfrac{a(1- r^{n})}{1-r}\:\:; r< 0 \right]$

$\implies S = \dfrac{8}{9}\left[\dfrac{10^{n+1} - 10-9n}{9}\right]$

$\implies S = \dfrac{8}{81}\left[10^{n+1} - 9n - 10 \right]$

$\implies S = \dfrac{8}{81}\left[10^{n} \cdot 10 - 9n - 10 \right]$

$\implies S = \dfrac{80}{81} (10^{n} – 1) – \dfrac{8}{9}n$

$\textbf{Shortcut:}$ We can check, by taking the values of $n = 1,2,3,\dots.$

So, the correct answer is $(D).$
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