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The sum of $n$ terms of the series $4+44+444+ \dots \dots$ is

1. $\frac{4}{81}\left[10^{n+1}-9n-1\right]$
2. $\frac{4}{81}\left[10^{n-1}-9n-1\right]$
3. $\frac{4}{81}\left[10^{n+1}-9n-10\right]$
4. $\frac{4}{81}\left[10^{n}-9n-10\right]$

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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

$S_n = \left(\frac{10^{n}-1}{9} \right)+ \left(\frac{10^{n-1}-1}{9} \right)+\left(\frac{10^{n-2}-1}{9} \right)+\ldots+\left(\frac{10^{1}-1}{9} \right)+(S_{0})$

$\implies S_n = \frac{1}{9}\left[\left(10^{n}-1 \right)+ \left(10^{n-1}-1\right )+\left(10^{n-2}-1\right )+\ldots+\left(10^{1}-1\right ) \right]$

$\implies S_n = \frac{1}{9}\left[10^{n}+ 10^{n-1}+10^{n-2}+\ldots+10^{1}- n.1 \right]$

$\implies S_n = \frac{1}{9}\left[10^{1}+ 10^{2}+10^{3}+\ldots+10^{n}- n \right]$

$\implies S_n = \frac{1}{9}\left[\left(\frac{10^{n}-1}{9}\right)- n \right] = \frac{1}{9\times 9}\left[10^{n+1}-10-9 n \right] = \frac{1}{81}\left[10^{n+1}-9n-10 \right]$

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 $n4 terms. So, let$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!)

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