1
harsh jindal
·2010-02-11 09:12:23
someone try this yaar............
341
Hari Shankar
·2010-02-11 22:13:18
Giveaway hint: The substitution a_n = \cot \theta gives you a_{n+1} = \cot \frac {\theta}{2}
1
harsh jindal
·2010-02-12 02:47:38
YES CORRECT,
F(n)= cot\left(\frac{\Pi }{2^{n+1}} \right)
1
Great Dreams
·2010-02-16 03:46:50
the question is wrong
i think , u meant
a_{n+1}=a_{n}+\sqrt{1+(a_{n})^2}
then u can square both sides
\left( a_{n+1}-a_n\right)^{2}=1+a_n^2 \\ \Rightarrow a_{n+1}^2-2a_n.a_{n+1}=1\\ \Rightarrow \frac{a_{n+1}^2-1}{2a_{n+1}}=a_n
this closely represents a famous trig formula [6][4]
1
harsh jindal
·2010-02-16 05:21:17
oops ,sorry, i missed this from question [4]