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duynhan1

[TEX]S = \frac12 + \frac{3}{2^2} +\frac{5}{2^3} ...+ \frac{2n-1}{2^n}[/TEX]
[TEX]\frac12 S = \frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4}+...+ \frac{2n-1}{2^{n+1}} [/TEX]

[TEX]S - \frac12 S = \frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3}+...+ \frac{2}{2^n} - \frac{2n-1}{2^{n+1} } [/TEX]

[TEX]\Leftrightarrow \frac12 S = \frac12 + ( \frac12 + \frac{1}{2^2} + ..+ \frac{1}{2^{n-1}}) - \frac{2n-1}{2^{n+1}} = \frac12 + \frac{\frac12 ( 1 - \frac{1}{2^{n-1})}}{1 - \frac12} - \frac{2n-1}{2^{n+1}}= \frac12 + 1 - \frac{1}{2^{n-1}}- \frac{2n-1}{2^{n+1}} [/TEX]
[TEX]\Leftrightarrow S = 3- \frac{2n+3}{2^{n}} [/TEX]
 
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