Question

The sum of the series $1+\frac{2}{11}+\frac{5}{{11}^2}+\frac{10}{{11}^3}+\frac{17}{{11}^4}+..$ is

• A. $\frac{154}{125}$
• B. $\frac{32}{25}$
• C. $\frac{363}{250}$
• D. $\frac{215}{175}$

Answer 1

The correct option is A $\frac{154}{125}$

$a=1+\frac{2}{11}+\frac{5}{{11}^2}+\frac{10}{{11}^3}+\frac{17}{{11}^4}\frac{a}{11}=\frac{1}{11}+\frac{2}{{11}^2}+\frac{5}{{11}^3}+\frac{10}{{11}^4}+....a-\frac{a}{11}=1+\frac{1}{11}+\frac{3}{{11}^2}+\frac{5}{{11}^3}+\frac{7}{{11}^4}+....\frac{10a}{11}=1+\frac{1}{11}+\frac{3}{{11}^2}+\frac{5}{{11}^3}+\frac{7}{{11}^4}+\frac{9}{{11}^5}+....\frac{10a}{{11}^2}=\frac{1}{11}+\frac{1}{{11}^2}+\frac{3}{{11}^3}+\frac{5}{{11}^4}+\frac{7}{{11}^5}+....\frac{10a}{11}-\frac{10a}{{11}^2}=1+0+\frac{2}{{11}^2}+\frac{2}{{11}^3}+\frac{2}{{11}^4}+\frac{2}{{11}^5}+....\frac{100a}{121}=1+\frac{2}{{11}^2}(1+\frac{1}{11}+\frac{1}{{11}^2}+....)\frac{100a}{121}=1+\frac{2}{{11}^2}\left(\frac{1}{1-\frac{1}{11}}\right)\frac{100a}{121}=1+\frac{2}{110}a=\frac{112}{110}\times \frac{121}{100}a=\frac{154}{125}1+\frac{2}{11}+\frac{5}{{11}^2}+\frac{10}{{11}^3}+\frac{17}{{11}^4}+....=\frac{154}{125}$