Discussion

Explanation:

In this special series, nth term can be written as:

T_n = $\frac{2^n-1}{2^n}$ = 1 - $\frac{1}{2^n}$

Hence, 
T₁ = 1 – 1/2
T₂ = 1 – 1/2²
T₃ = 1 – 1/2³
…
T₁₁ = 1 – 1/2¹¹

S₁₁ = 20 – $\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\frac{1}{2^{11}}\right)$
      = 20 - $\frac{{\displaystyle \frac{1}{2}}\left[{\left({\displaystyle \frac{1}{2}}\right)}^{11}-1\right]}{{\displaystyle \frac{1}{2}}-1}$
      = 20 + 1/2¹¹  – 1
      = 19 + 1/2¹¹ 

Hence, option (b).

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