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The sum of the series $3 + 33 + 333 + ... + n$ terms is
$\frac{1}{{27}}({10^{n + 1}} + 9n - 28)$
$\frac{1}{{27}}({10^{n + 1}} - 9n - 10)$
$\frac{1}{{27}}({10^{n + 1}} + 10n - 9)$
None of these
Solution
(b) Series $3 + 33 + 333 +………+ n$ terms
Given series can be written as,
$ = \frac{1}{3}[9 + 99 + 999 + …….. + n\,\,{\rm{terms]}}$
$ = \frac{1}{3}\left[ {(10 – 1) + ({{10}^2} – 1) + ({{10}^3} – 1) + …. + n\,{\rm{terms}}} \right]$
$ = \frac{1}{3}\left[ {10 + {{10}^2} + …. + {{10}^n}} \right]$$ – \frac{1}{3}\left[ {1 + 1 + 1 + …. + n\,{\rm{terms}}} \right]$
$ = \frac{1}{3}\,.\,\frac{{10\,({{10}^n} – 1)}}{{10 – 1}} – \frac{1}{3}.n\,$ $ = \frac{1}{3}\left[ {\frac{{{{10}^{n + 1}} – 10}}{9} – n} \right]$
$ = \frac{1}{3}\,\left[ {\frac{{{{10}^{n\, + \,1}} – 9n – 10}}{9}} \right]$ $ = \frac{1}{{27}}[{10^{n\, + \,1}} – 9n – 10]$.
