A) \[\frac{121}{10}\]
B) \[\frac{441}{100}\]
C) \[100\]
D) \[110\]
Correct Answer: C
Solution :
Let \[K=1+2\left( \frac{11}{10} \right)+3{{\left( \frac{11}{10} \right)}^{2}}+...+10{{\left( \frac{11}{10} \right)}^{9}}\]?(1) Also, \[\frac{11K}{10}=\frac{11}{10}+2{{\left( \frac{11}{10} \right)}^{2}}+...3{{\left( \frac{11}{10} \right)}^{9}}+10{{\left( \frac{11}{10} \right)}^{10}}\] (1)-(2) gives \[-\frac{k}{10}=1+\left( \frac{11}{10} \right)+{{\left( \frac{11}{10} \right)}^{2}}+...+{{\left( \frac{11}{10} \right)}^{9}}-10\cdot {{\left( \frac{11}{10} \right)}^{10}}\]\[\Rightarrow \]\[-\frac{k}{10}=\frac{\left( {{\left( \frac{11}{10} \right)}^{10}}-1 \right)}{\frac{11}{10}-1}-10\cdot {{\left( \frac{11}{10} \right)}^{10}}\] \[\Rightarrow \]\[-\frac{k}{10}=10.{{\left( \frac{11}{10} \right)}^{9}}-10-10{{\left( \frac{11}{10} \right)}^{10}}\Rightarrow k=100\]
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