JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q19

← JEE Main 2022 · 28 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 19 of 90

Consider the sequence $a_1,a_2,a_3,\ldots$ such that $a_1=1$ , $a_2=2$ and $a_{n+2}=\dfrac2{a_{n+1}}+a_n$ for $n=1,2,3,\ldots$ . If

$\left(\frac{a_1+\frac1{a_2}}{a_3}\right)\cdot\left(\frac{a_2+\frac1{a_3}}{a_4}\right)\cdot\left(\frac{a_3+\frac1{a_4}}{a_5}\right)\cdots\left(\frac{a_{30}+\frac1{a_{31}}}{a_{32}}\right)=2^\alpha\left({}^{61}C_{31}\right),$

then $\alpha$ is equal to :

-30

-31

-60

-61

Answer and worked solution

Official answer: C

Telescoping relation. $a_{n+2}a_{n+1}=2+a_na_{n+1}$ , so $a_na_{n+1}=2n$ and $\frac{a_n+1/a_{n+1}}{a_{n+2}}=\frac{a_na_{n+1}+1}{a_{n+1}a_{n+2}}=\frac{2n+1}{2n+2}$ .

Product $\prod_{n=1}^{30}\frac{2n+1}{2n+2}=\frac{61!}{2^{60}(30!)(31!)}=2^{-60}\,{}^{61}C_{31}$ .

$\alpha=-60$ . (C)

Topics:Special series sums Properties of binomial coefficients

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JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 19