The sum of n terms of the series $4+44+444+ \ldots$ is
Final sum $= 4(1+11+111+1111+\ldots)$
Let $S_n = \underbrace{1+11+111+1111+\ldots +1111}_{n \text{ times}}$
∴ $S_0 = 0$ and $S_1 = 1$ and $S_2 = 1+11 = 12.$
$S_n = (10^0)+ (10^1+1)+(10^2+11)+(10^3+111)+\ldots+ \left(10^{n-1}+\underbrace{111\ldots 1}_{n-1 \text{ times}}\right)$
$\implies S_n = ({10^0+10^1+10^2+10^3+\ldots10^{n-1}} )+ \left({1+11+111+1111+\ldots+\underbrace{111\ldots 1}_{n-1 \text{ times}} }\right)$
- $\implies S_n = \left (\frac{10^{n}-1}{9} \right)+ (S_{n-1})\quad \to (1)$
Recursively substituting, we get
$S_n = \left(\frac{10^{n}-1}{9} \right)+ \left(\frac{10^{n-1}-1}{9} \right)+\left(\frac{10^{n-2}-1}{9} \right)+\ldots+\left(\frac{10^{1}-1}{9} \right)+(S_{0})$
$\implies S_n = \frac{1}{9}\left[\left(10^{n}-1 \right)+ \left(10^{n-1}-1\right )+\left(10^{n-2}-1\right )+\ldots+\left(10^{1}-1\right ) \right]$
$\implies S_n = \frac{1}{9}\left[10^{n}+ 10^{n-1}+10^{n-2}+\ldots+10^{1}- n.1 \right]$
$\implies S_n = \frac{1}{9}\left[10^{1}+ 10^{2}+10^{3}+\ldots+10^{n}- n \right]$
$\implies S_n = \frac{1}{9}\left[10\left(\frac{10^{n}-1}{9}\right)- n \right] = \frac{1}{9\times 9}\left[10^{n+1}-10-9 n \right] = \frac{1}{81}\left[10^{n+1}-9n-10 \right]$
Final SUM $= 4 S_n = \frac{4}{81}\left[10^{n+1}-10-9 n \right] = \frac{4}{81}\left[10^{n+1}-9n-10 \right]$
Shortcut :-
in the question, they asked for sum of $n$ terms.
So, let $n=3$
$\implies$ Required sum $= 4+44+444 = 492$
Put in each option $n=3,$ it matches only for Option C.
(If more than one option is matched, then take $n=4,n=5$ until you get only one option is matched!)
Correct Answer: Option C.