A) \[\frac{10}{81}\,\,({{10}^{n}}-1)-\frac{n}{81}\]
B) \[\frac{10}{81}\,\,({{10}^{n}}-1)-\frac{n}{9}\]
C) \[\frac{10}{81}\,\,({{10}^{n}}-1)+\frac{n}{9}\]
D) \[\frac{10}{81}\,\,({{10}^{n}}-1)+\frac{n}{81}\]
Correct Answer: B
Solution :
Given, series \[1+11+111+....+n\] terms \[=\frac{1}{9}\,[9+99+999+....+n\,terms]\] \[=\frac{1}{9}[(10-1)+(100-1)+(1000-1)\] \[+....+n\,terms]\] \[=\frac{1}{9}[10+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{n}}-n]\] \[=\frac{1}{9}\left[ 10.\left( \frac{{{10}^{n}}-1}{10-1} \right)-n \right]\] \[=\frac{1}{9}\left[ \frac{10}{9}({{10}^{n}}-1)-n \right]\] \[=\frac{10}{81}\,({{10}^{n}}-1)-\frac{n}{9}\]
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