The sum of the series 3 + 33 + 333 + dots + n | vedclass

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(B) Let the sum be $S_n = 3 + 33 + 333 + \dots + n$ terms.We can write $S_n = 3(1 + 11 + 111 + \dots + n 	ext{ terms})$.Multiply and divide by $9$:$S

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

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(B) Let the sum be $S_n = 3 + 33 + 333 + \dots + n$ terms.We can write $S_n = 3(1 + 11 + 111 + \dots + n 	ext{ terms})$.Multiply and divide by $9$:$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n 	ext{ terms})$$S_n = \frac{1}{3}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n 	ext{ terms})]$The first part is a geometric progression with $a = 10$,$r = 10$,and $n$ terms. The sum is $\frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.$S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{9} - n ight]$$S_n = \frac{1}{3} \left[ \frac{10^{n+1} - 10 - 9n}{9} ight]$$S_n = \frac{1}{27}(10^{n+1} - 9n - 10)$.

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

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