The sum of the first 10 terms of the series 9 | vedclass

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(B) The given series is $S_{10} = 9 + 99 + 999 + \ldots$ up to $10$ terms.This can be written as $S_{10} = (10-1) + (10^2-1) + (10^3-1) + \ldots + (10

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$\frac{100}{9}(10^9-1)$

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(B) The given series is $S_{10} = 9 + 99 + 999 + \ldots$ up to $10$ terms.This can be written as $S_{10} = (10-1) + (10^2-1) + (10^3-1) + \ldots + (10^{10}-1)$.$S_{10} = (10 + 10^2 + 10^3 + \ldots + 10^{10}) - (1 + 1 + 1 + \ldots + 1 	ext{ (10 times)})$.The first part is a geometric progression with $a=10$,$r=10$,and $n=10$.Sum of $GP$ $= a\frac{r^n-1}{r-1} = 10\frac{10^{10}-1}{10-1} = \frac{10}{9}(10^{10}-1)$.Subtracting the sum of $1$s (which is $10$):$S_{10} = \frac{10}{9}(10^{10}-1) - 10 = \frac{10(10^{10}-1) - 90}{9} = \frac{10^{11}-10-90}{9} = \frac{10^{11}-100}{9} = \frac{100}{9}(10^9-1)$.

The sum of the first $10$ terms of the series $9+99+999+\ldots$ is

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