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what is the next 3 terms of.......tanx=x + 'x'qube by3 +2.'x'raised to a power5 by15OR X+X^3/3+2X^5/15+ NEXT 3 TERS WHAT??????

Praveen Kumar beniwal , 13 Years ago
Grade
anser 1 Answers
Pramod J AskiitiansExpert-IIT-B

Last Activity: 13 Years ago

Dear student,

\tan x = \sum^{\infin}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1} = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \cdots\text{ for }|x| < \frac{\pi}{2}\!

 

where B is a bernoulli series

B_m(n)=\sum_{k=0}^m\sum_{v=0}^k(-1)^v\binom kv\frac{\left( n+v\right) ^m}{k+1} ,

 

fourth term of the series is 17x^7/315 other two 2 can be solved by the above formula

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