1YES CORRECT,
F(n)= cot\left(\frac{\Pi }{2^{n+1}} \right)
TRY THIS QUESTION:
A recurrence satisfies
a_{n+1}=a_{n}+\sqrt{1+a_{n}}
then
Obtain an explicit formula for a_{n} i.e. find a function f(n) such that
a_{n}=F(n)
(given that a_{0}=0 )
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5 Answers
341Giveaway hint: The substitution a_n = \cot \theta gives you a_{n+1} = \cot \frac {\theta}{2}
1the 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]