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\[ \large \sum _{ n=0 }^{ 2015 } { \frac { 1 }{ { 2 }^{ n }+\sqrt { { 2 }^{ 2015 } } } } = \, ? \]

Note by Siva Prasad 2 years, 1 month ago

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\(\begin{equation} \begin{split} & = \displaystyle \sum^{2015}_{n=0}\frac{1}{2^n+\sqrt{ 2^{2015} }} \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }}+\frac{1}{2^{2015-n} + \sqrt{ 2^{2015}} } \right) \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }} + \frac{2^n}{2^{2015}+2^n \sqrt{ 2^{2015} } } \right) \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }} + \frac{2^n}{ \sqrt{2^{2015}}( \sqrt{2^{2015}}+2^n)} \right) \\ & = \displaystyle \sum^{1007}_{n=0} \left( \frac{ \sqrt{2^{2015}}+2^n}{ \sqrt{2^{2015}}( \sqrt{2^{2015}}+2^n)} \right) \\ & = \displaystyle \sum^{1007}_{n=0} \frac{1}{\sqrt{2^{2015}}} \\ & = \frac{1008}{\sqrt{2^{2015}}} \\ & = \boxed{ \frac{63}{\sqrt{2^{2007}}}} \end{split} \end{equation} \)

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Nice solution!! Upvoted.

Use \(\sum^{b}_{r=a}f(r)=\sum_{r=a}^{b}f(a+b-r)\)

Add both of them , simplify you will get \(\Large{2S= \frac{1}{2^{\frac{2015}{2}}}\displaystyle \sum^{2015}_{n=0} 1 \Rightarrow 2S=\frac{1}{2^{\frac{2015}{2}}}\times 2016}\)

What's \(f(r)\) and \(S\) ?

\(\large{S}\) is the given summation.

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TopNewest\(\begin{equation} \begin{split} & = \displaystyle \sum^{2015}_{n=0}\frac{1}{2^n+\sqrt{ 2^{2015} }} \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }}+\frac{1}{2^{2015-n} + \sqrt{ 2^{2015}} } \right) \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }} + \frac{2^n}{2^{2015}+2^n \sqrt{ 2^{2015} } } \right) \\ & = \displaystyle \sum^{1007}_{n=0}\left( \frac{1}{2^n + \sqrt{ 2^{2015} }} + \frac{2^n}{ \sqrt{2^{2015}}( \sqrt{2^{2015}}+2^n)} \right) \\ & = \displaystyle \sum^{1007}_{n=0} \left( \frac{ \sqrt{2^{2015}}+2^n}{ \sqrt{2^{2015}}( \sqrt{2^{2015}}+2^n)} \right) \\ & = \displaystyle \sum^{1007}_{n=0} \frac{1}{\sqrt{2^{2015}}} \\ & = \frac{1008}{\sqrt{2^{2015}}} \\ & = \boxed{ \frac{63}{\sqrt{2^{2007}}}} \end{split} \end{equation} \)

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Nice solution!! Upvoted.

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Use \(\sum^{b}_{r=a}f(r)=\sum_{r=a}^{b}f(a+b-r)\)

Add both of them , simplify you will get \(\Large{2S= \frac{1}{2^{\frac{2015}{2}}}\displaystyle \sum^{2015}_{n=0} 1 \Rightarrow 2S=\frac{1}{2^{\frac{2015}{2}}}\times 2016}\)

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What's \(f(r)\) and \(S\) ?

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\(\large{S}\) is the given summation.

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