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∑n=0201512n+22015= ? \large \sum _{ n=0 }^{ 2015 } { \frac { 1 }{ { 2 }^{ n }+\sqrt { { 2 }^{ 2015 } } } } = \, ?

#Algebra

Note by Siva Prasad
5 years, 5 months ago

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Comments

=∑n=0201512n+22015=∑n=01007(12n+22015+122015−n+22015)=∑n=01007(12n+22015+2n22015+2n22015)=∑n=01007(12n+22015+2n22015(22015+2n))=∑n=01007(22015+2n22015(22015+2n))=∑n=01007122015=100822015=6322007 \begin{aligned} & = \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{aligned}

Akshat Sharda - 5 years, 5 months ago

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Nice solution!! Upvoted.

Anshuman Singh Bais - 5 years, 5 months ago

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

Add both of them , simplify you will get 2S=1220152∑n=020151⇒2S=1220152×2016\Large{2S= \frac{1}{2^{\frac{2015}{2}}}\displaystyle \sum^{2015}_{n=0} 1 \Rightarrow 2S=\frac{1}{2^{\frac{2015}{2}}}\times 2016}

Tanishq Varshney - 5 years, 5 months ago

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What's f(r)f(r) and SS ?

Akshat Sharda - 5 years, 5 months ago

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S\large{S} is the given summation.

Tanishq Varshney - 5 years, 5 months ago
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