Find the sum of the series 6 + 66 + 666 + dot | vedclass

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(B) Let $S_n = 6 + 66 + 666 + \dots + n 	ext{ terms}$.$S_n = \frac{6}{9} (9 + 99 + 999 + \dots + n 	ext{ terms})$.$S_n = \frac{2}{3} [(10 - 1) + (10

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$\frac{2(10^{n+1} - 9n - 10)}{27}$

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(B) Let $S_n = 6 + 66 + 666 + \dots + n 	ext{ terms}$.$S_n = \frac{6}{9} (9 + 99 + 999 + \dots + n 	ext{ terms})$.$S_n = \frac{2}{3} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$.$S_n = \frac{2}{3} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n 	ext{ times})]$.Using the sum of a geometric progression formula $S = \frac{a(r^n - 1)}{r - 1}$:$S_n = \frac{2}{3} [\frac{10(10^n - 1)}{10 - 1} - n]$.$S_n = \frac{2}{3} [\frac{10(10^n - 1)}{9} - n]$.$S_n = \frac{2}{3} [\frac{10^{n+1} - 10 - 9n}{9}]$.$S_n = \frac{2(10^{n+1} - 9n - 10)}{27}$.

Find the sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms.

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