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2013 ∑ tan(x/2^(n)).sec(x/2^(n-1)) = tan(x/2^(a)) - tan(x/2^(b)) n=1

2013
  ∑    tan(x/2^(n)).sec(x/2^(n-1)) = tan(x/2^(a)) - tan(x/2^(b))
n=1

Grade:11

1 Answers

jagdish singh singh
173 Points
6 years ago
\hspace{-0.5 cm }\sum^{2013}_{n=1}\tan\left(\frac{x}{2^n}\right)\cdot \sec\left(\frac{x}{2^{n-1}}\right)=\sum^{2013}_{n=1}\frac{\sin (\frac{x}{2^n})}{\cos \left(\frac{x}{2^n}\right)\cdot \cos \left(\frac{x}{2^{n-1}}\right)}\\\\\\ \sum^{2013}_{n=1}\frac{\sin \left[\frac{x}{2^{n-1}}-\frac{x}{2^n}\right]}{\cos \left(\frac{x}{2^n}\right)\cdot \cos \left(\frac{x}{2^{n-1}}\right)} = \sum^{2013}_{n=1}\left[\tan \left(\frac{x}{2^{n-1}}\right)-\tan\left(\frac{x}{2^n}\right)\right]$\\\\\\ Now expanding Summation, We get Sum equal to\\\\$=\tan \left(\frac{x}{2^{0}}\right)-\tan \left(\frac{x}{2^{2013}}\right)$

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