Animated Solution for Mathematics - Sequence and Series: Consider the sequence $a_1,a_2,a_3,\dots$ such that $a_1=1,a_2=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_n$ for $n=1,2,3,\dots$. If $\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right)\cdot \left(\frac{a_2+\frac{1}{a_3}}{a_4}\right)\cdot \left(\frac{a_3+\frac{1}{a_4}}{a_5}\right)\dots \left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha }\left(\genfrac{}{}{0ex}{}{61}{31}\right)$, then $\alpha$ is equal to :

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of fractions and indices.

You see a recurrence relation like $a_{n+2}=\frac{2}{a_{n+1}}+a_n$ and your instinct might be to panic. But in the world of JEE Advanced, panic is the enemy. We are going to use logic, pattern recognition, and a bit of algebraic elegance to turn this chaos into order.

The Linearization

The problem gives us $a_1=1$ and $a_2=2$. The recurrence $a_{n+2}=\frac{2}{a_{n+1}}+a_n$ is the heart of the beast.

That denominator, $a_{n+1}$, is the source of all our trouble. In mathematics, when you see a denominator that complicates your life, your first instinct should be to clear it. Let us multiply the entire equation by $a_{n+1}$:

Look at that! The fraction is gone. The equation is now linear in terms of products. This is the moment where the problem shifts from 'impossible' to 'solvable'.

The Hidden Arithmetic Progression

Now, let us define a new sequence, $b_n=a_n{a}_{n+1}$. If we substitute this into our linearized equation, we get:

This is the definition of an Arithmetic Progression (A.P.) with a common difference $d=2$. We know $b_1=a_1{a}_2=(1)(2)=2$.

Using the standard formula for the $n$-th term of an A.P., $b_n=b_1+(n-1)d$, we find:

So, we have discovered that $a_n{a}_{n+1}=2n$. This is a massive simplification. We have reduced a complex recurrence to a simple linear product.

The Product Transformation

Now, let us look at the product we need to evaluate. Let $T_n$ be the general term of the product:

If we simplify the numerator by taking the common denominator $a_{n+1}$, we get:

Using our definition $b_n=a_n{a}_{n+1}$, this becomes:

This is beautiful. The entire product $P$ is just the product of these $T_n$ terms from $n=1$ to $30$.

The Grand Finale

We have $P={\prod }_{n=1}^{30}\frac{2n+1}{2(n+1)}$. We can pull out the factor of $\frac{1}{2}$ thirty times:

The numerator is the product of odd numbers, and the denominator is $31!$. To make the numerator a full factorial, we multiply and divide by the product of even numbers $(2\cdot 4\cdot 6\dots 60)$:

The product of even numbers is $2^{30}\cdot 30!$. Substituting this back:

Comparing this to $2^{\alpha }\left(\genfrac{}{}{0ex}{}{61}{31}\right)$, we find $\alpha =-60$. You have conquered the problem. Remember, the math is not just about calculation; it is about seeing the structure beneath the surface.