JEE Main 2022 — Sequences And Series Question with Solution

$$a_{n+2}=\frac{2}{a_{n+1}}+a_n$$

$$\Rightarrow a_n{a}_{n+1}+1=a_{n+1}{a}_{n+2}-1$$

$$\Rightarrow a_{n+2}{a}_{n+1}-a_n.a_{n+1}=2$$

For

\matrix{ {n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr {n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \cr {n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \cr {} & {\matrix{ . \cr . \cr . \cr . \cr } } \cr {n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \cr }

Now,

$$\frac{(a_1{a}_2+1)}{a_2{a}_3}.\frac{(a_2{a}_3+1)}{a_3{a}_4}.\frac{(a_3{a}_4+1)}{a_4{a}_5}.......\frac{(a_{30}{a}_{31}+1)}{a_{31}{a}_{32}}$$

$$=\frac{3}{4}.\frac{5}{6}.\frac{7}{8}......\frac{61}{62}$$

$$=2^{-60}\left({}^{61}{C}_{31}\right)$$