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Question Number 192149 by Shrinava last updated on 09/May/23

Find :

\frac{1}{2} + \frac{3}{2^{3}} + \frac{5}{2^{5}} + \frac{7}{2^{7}} + \dots

Answered by aleks041103 last updated on 09/May/23

\underset{k = 1}{\sum^{\infty}} (2 k - 1) x^{2 k - 1} = x {(\underset{k = 1}{\sum^{\infty}} x^{2 k - 1})}^{'} =

= x {(x \underset{k = 1}{\sum^{\infty}} x^{2 (k - 1)})}^{'} = x {(x \frac{1}{1 - x^{2}})}^{'} =

= x \frac{d}{d x} (\frac{x}{1 - x^{2}}) = x \frac{(1 - x^{2}) - (- 2 x) x}{1 - x^{2}} =

= x \frac{1 + x^{2}}{1 - x^{2}}

\Rightarrow \underset{k = 1}{\sum^{\infty}} \frac{2 k - 1}{2^{2 k - 1}} = {(x \frac{1 + x^{2}}{1 - x^{2}})}_{x = 1 / 2} = \frac{1}{2} \frac{1 + \frac{1}{4}}{1 - \frac{1}{4}} =

= \frac{1}{2} \frac{5}{3}

\Rightarrow \frac{1}{2} + \frac{3}{2^{3}} + \frac{5}{2^{5}} + \dots = \frac{5}{6}

Commented by AST last updated on 09/May/23

\frac{d}{d x} (\frac{x}{1 - x^{2}}) = \frac{1 + x^{2}}{{(+ x^{2} - 1)}^{2}}

\Rightarrow x \frac{d}{d x} (\frac{x}{1 - x^{2}}) = \frac{x (1 + x^{2})}{{(+ x^{2} - 1)}^{2}}

\Rightarrow F o r x = \frac{1}{2}; w e g e t \frac{10}{9}

Commented by BaliramKumar last updated on 09/May/23

\frac{1}{2} + \frac{3}{2^{3}} + \frac{5}{2^{5}} = \frac{33}{32} > 1

Answered by mehdee42 last updated on 09/May/23

w e w i l l c o n s i d e r; A = x + 3 x^{3} + 5 x^{5} + \dots; 0 < x < 1

\Rightarrow x^{2} A = x^{3} + 3 x^{5} + 5 x^{7} + \dots

\Rightarrow A - {A x}^{2} = x + 2 (x^{3} + x^{5} + x^{7} + \dots)

(1 - x^{2}) A = x + \frac{2 x^{3}}{1 - x^{2}} \Rightarrow A = \frac{x + x^{3}}{{(1 - x^{2})}^{2}}

n o w r e g a r d i n g t h e q u e s t i o n, i f “ x = \frac{1}{2}^{″} w e w i l l h a v e

a n s w e r = \frac{10}{9} ✓

Answered by universe last updated on 09/May/23

s = \frac{1}{2} + \frac{3}{2^{3}} + \frac{5}{2^{5}} + \frac{7}{2^{7}} + \dots

\frac{1}{4} s = \frac{1}{2^{3}} + \frac{3}{2^{5}} + \frac{5}{2^{7}} + \dots

\frac{3}{4} s = \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{4}} + \frac{1}{2^{6}} + \dots

\frac{3}{4} s = \frac{1}{2} + \frac{1 / 2^{2}}{1 - 1 / 2^{2}}

\frac{3}{4} s = \frac{1}{2} + \frac{1}{3}

s = \frac{5}{6} \times \frac{4}{3} = \frac{20}{18}

s = \frac{10}{9}

Answered by York12 last updated on 09/May/23

{t h a t}^{'} s A G P

Answered by manxsol last updated on 11/May/23

t_{k} = q \frac{2 k - 1}{2^{2 k - 1}}

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