The-sum-of-the-first-n-terms-of-the-series-1-2-3-4-7-8-15-16-is-equal-to-

Question Number 11246 by 786786AM last updated on 18/Mar/17

The sum of the first n terms of the series

\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots is equal to

Answered by FilupS last updated on 18/Mar/17

\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots = \underset{k = 1}{\sum^{n}} \frac{2^{k} - 1}{2^{k}}

\underset{k = 1}{\sum^{n}} \frac{2^{k} - 1}{2^{k}} = \underset{k = 1}{\sum^{n}} 1 - \frac{1}{2^{k}}

= \underset{k = 1}{\sum^{n}} 1 - \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}}

= (\underset{k = 1}{\sum^{n}} 1) - S S = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}}

\underset{k = 1}{\sum^{n}} \frac{2^{k} - 1}{2^{k}} = n - S (1)

S = \underset{k = 1}{\sum^{n}} \frac{1}{2^{k}}

S = \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots + \frac{1}{2^{n}}

S = \frac{1}{2^{1}} (1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n - 1}})

S = \frac{1}{2} (1 + S - \frac{1}{2^{n}})

S = \frac{1}{2} + \frac{1}{2} S - \frac{1}{2^{n + 1}}

S - \frac{1}{2} S = \frac{1}{2} - \frac{1}{2^{n + 1}}

S (1 - \frac{1}{2}) = \frac{2^{n + 1} - 2}{2^{n + 2}}

S (1 - \frac{1}{2}) = \frac{2 (2^{n} - 1)}{2^{n + 2}}

S (1 - \frac{1}{2}) = \frac{(2^{n} - 1)}{2^{n + 1}}

S = \frac{(2^{n} - 1)}{2^{n + 1} (1 - \frac{1}{2})}

S = \frac{(2^{n} - 1)}{2^{n + 1} - \frac{1}{2} 2^{n + 1}}

S = \frac{(2^{n} - 1)}{2^{n + 1} - 2^{n}}

S = \frac{(2^{n} - 1)}{2^{n} (2^{1} - 1)}

S = \frac{2^{n} - 1}{2^{n}} (2)

sub (2) \to (1)

\underset{k = 1}{\sum^{n}} \frac{2^{k} - 1}{2^{k}} = n - S

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

= n - (\frac{2^{n}}{2^{n}} - \frac{1}{2^{n}})

= n - (1 - \frac{1}{2^{n}})

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

∴ \underset{k = 1}{\sum^{n}} \frac{2^{k} - 1}{2^{k}} = \frac{1}{2} + \frac{3}{4} + \dots = n + \frac{1}{2^{n}} - 1

Answered by sandy_suhendra last updated on 18/Mar/17

= (1 - \frac{1}{2}) + (1 - \frac{1}{4}) + (1 - \frac{1}{8}) + \dots

= (1 + 1 + 1 + \dots) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots)

= n - \frac{\frac{1}{2} [1 - {(\frac{1}{2})}^{n}]}{1 - \frac{1}{2}}

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

Commented by FilupS last updated on 18/Mar/17

I did my proof without the formula : P

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