A) \[\frac{121}{10}\]
B) \[\frac{441}{100}\]
C) \[100\]
D) \[110\]
Correct Answer: C
Solution :
| We have,\[{{(10)}^{9}}k={{10}^{9}}+2\,(11){{(10)}^{8}}+3\,{{(11)}^{2}}{{(10)}^{7}}+...+10\,{{(11)}^{9}}\]\[\Rightarrow \]\[k=1+2\left( \frac{11}{10} \right)+3{{\left( \frac{11}{10} \right)}^{2}}+...+10{{\left( \frac{11}{10} \right)}^{9}}\] |
| \[\Rightarrow \] \[\frac{11}{10}k=\frac{11}{10}+2{{\left( \frac{11}{10} \right)}^{2}}+...9{{\left( \frac{11}{10} \right)}^{9}}+{{\left( \frac{11}{10} \right)}^{10}}10-\frac{k}{10}\] |
| \[=1+\frac{11}{10}+{{\left( \frac{11}{10} \right)}^{2}}+...{{\left( \frac{11}{10} \right)}^{9}}-{{\left( \frac{11}{10} \right)}^{10}}10\]\[-\frac{k}{10}=\frac{{{\left( \frac{11}{10} \right)}^{10}}-1}{\frac{11}{10}-1}-{{\left( \frac{11}{10} \right)}^{10}}10\] |
| \[\therefore \] \[k=100\] |
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