MCQ
$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to
- A$\frac {2}{9}$
- ✓$\frac {4}{9}$
- C$\frac {4}{3}$
- D$\frac {2}{3}$
$\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} \left( {2 - \frac{{n + 1}}{{{2^{n - 1}}}}} \right)$
$\sum\limits_{n = 1}^\infty {\frac{1}{{{2^{n - 1}}}} - \frac{{n + 1}}{{{2^{2n - 1}}}}} = \frac{4}{9}$
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