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

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Question

(b) Series $3 + 33 + 333 +&hellip;......+ n$ terms

Given series can be written as,

$ = \frac{1}{3}[9 + 99 + 999 + ........ + n\,\,{\rm{terms]}}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Series $3 + 33 + 333 +&hellip;......+ n$ 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]$.

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

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