• question_answer If ${{(10)}^{9}}+2{{(11)}^{1}}{{(10)}^{8}}+3{{(11)}^{2}}{{(10)}^{7}}+...+10$${{(11)}^{9}}=k{{(10)}^{9}},$then k is equal to :   JEE Main  Solved  Paper-2014 A) $\frac{121}{10}$                                             B) $\frac{441}{100}$ C) $100$                  D) $110$

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$