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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
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
    $\frac {2}{3}$
✓

Answer

Correct option: B.
$\frac {4}{9}$
b
$\sum\limits_{n = 1}^\infty  {\frac{1}{{{2^n}}}} \left( {\frac{1}{{{2^1}}} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + ...... + \frac{{n - 1}}{{{2^{n - 1}}}}} \right)$

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{2^n}}}} \left( {2 - \frac{{n + 1}}{{{2^{n - 1}}}}} \right)$

$\sum\limits_{n = 1}^\infty  {\frac{1}{{{2^{n - 1}}}} - \frac{{n + 1}}{{{2^{2n - 1}}}}}  = \frac{4}{9}$

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