Question

Let $a_0=\frac{5}{2}$ and $a_k=a_{k-1}^2-2fork\ge 1,$ then the value of $\stackrel{\infty }{\underset{k=0}{\Pi }}(1-\frac{1}{a_k})$ is

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{7}$
• D. $\frac{4}{7}$

Answer 1

The correct option is C

$\frac{3}{7}$

$a_0=\frac{5}{2}=\frac{2^2+1}{2}{a}_1=a_0^2-2=\frac{17}{4}=\frac{2^4+1}{2^2}Let{a}_n=\frac{x^4+1}{x^2};x=2^{2^{n-1}}\therefore \stackrel{\infty }{\underset{k=0}{\pi }}(1-\frac{1}{a_k})=\underset{n\to \infty }{lim}\stackrel{n}{\underset{k=0}{\Pi }}\left(\frac{a_k-1}{a_k}\right)=\underset{n\to \infty }{lim}\frac{a_0-1}{a_0}.\frac{a_1-1}{a_1}.....\frac{a_n-1}{a_n}=\underset{n\to \infty }{lim}\frac{2^2-2+1}{2^2+1}.\frac{4^2-4+1}{4^2+1}.......\left(\frac{x^2-x+1}{x^2+1}\right)=\underset{n\to \infty }{lim}\frac{2^2-1}{2^2+2+1}\times \frac{x^4+x^2+1}{x^4-1}=\frac{3}{7}$