Sick-quences!! (2)

first of, by GP,
$1+\dfrac{1}{2}+...\dfrac{1}{2^n}=\dfrac{1(1-\dfrac{1}{2^{n+1}}}{1-\dfrac{1}{2}}=2- \dfrac{1}{2^n}$ now, writing in summation form, $\sum_{n=0}^\infty (\dfrac{1}{11^n}(2- \dfrac{1}{2^n}))= \sum_{n=0}^\infty (\dfrac{2}{11^n}- \dfrac{1}{22^n})$ now write it as $2\sum_{n=0}^\infty (\dfrac{1}{11^n})-\sum_{n=0}^\infty \dfrac{1}{22^n}$ let $S_1 =1+\dfrac{1}{11}+\dfrac{1}{11^2}.....$ $S_1=1+\dfrac{1}{11}S_1\longrightarrow S_1=\dfrac{11}{10}$ let $S_2 =1+\dfrac{1}{22}+\dfrac{1}{22^2}....$ $S_2=1+\dfrac{1}{22}S_2\longrightarrow S_2=\dfrac{22}{21}$ now put these values, $2(\dfrac{11}{10})-\dfrac{22}{21}=\dfrac{121}{105}$ and $121-105=\boxed{16}$

You can simply multiply by 1/11 and shift one term and subtract them.then again you repeat this process.you will automatically get a geometric progression and apply the sum of infinite Geometric progression and after doing little bit manipulations you will get 121/105 as answer.