Consider the sequence a_1,a_2,a_3,… such that a_1=1,a_2=2 and a_n+2= 2 a_n+1+a_n for n=1,2,3,… If ( a_1+ 1 a_2 a_3)· ( a_2+ 1 a_3 a_4)· ( a_3+ 1 a_4…

Given ${a}_{n+2}=\frac{2}{{a}_{n+1}}+{a}_{n}$an+2​=an+1​2​+an​

$$\Rightarrow {a}_{n+2}{a}_{n+1}-{a}_{n}{a}_{n+1}=2$$⇒an+2​an+1​−an​an+1​=2

Now, ${T}_{r}={a}_{r+1}{a}_{r}$Tr​=ar+1​ar​ is an A.P. with common difference $2$2

Now, ${T}_{1}={a}_{1}{a}_{2}=2,{T}_{2}={a}_{2}{a}_{3}=4,....,{T}_{r}=2r$T1​=a1​a2​=2,T2​=a2​a3​=4,....,Tr​=2r

So $\displaystyle (\frac{{a}_{1}+\frac{1}{{a}_{2}}}{{a}_{3}})\cdot (\frac{{a}_{2}+\frac{1}{{a}_{3}}}{{a}_{4}})\cdots (\frac{{a}_{30}+\frac{1}{{a}_{31}}}{{a}_{32}})=\prod _{i=1}^{30}(\frac{{a}_{i}+\frac{1}{{a}_{i+1}}}{{a}_{i+2}})$(a3​a1​+a2​1​​)⋅(a4​a2​+a3​1​​)⋯(a32​a30​+a31​1​​)=i=1∏30​(ai+2​ai​+ai+1​1​​)

$$=\prod _{i=1}^{30}(\frac{{a}_{i}{a}_{i+1}+1}{{a}_{i+1}{a}_{i+2}})=\prod _{i=1}^{30}\frac{{T}_{r}+1}{{T}_{r+1}}$$=i=1∏30​(ai+1​ai+2​ai​ai+1​+1​)=i=1∏30​Tr+1​Tr​+1​

$$=\prod _{i=1}^{30}(\frac{2r+1}{2r+2})=\frac{3\cdot 5\cdot 7\cdots 61}{4\cdot 6\cdot 8\cdots 62}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdots 61\cdot 62}{2{(4\cdot 6\cdot 8\cdots 62)}^{2}}$$=i=1∏30​(2r+22r+1​)=4⋅6⋅8⋯623⋅5⋅7⋯61​=2(4⋅6⋅8⋯62)21⋅2⋅3⋅4⋅5⋅6⋅7⋯61⋅62​

$$=\frac{62!}{2{(4\cdot 6\cdot 8\cdots 62)}^{2}}=\frac{62!}{{2}^{61}\cdot {(31!)}^{2}}=\frac{62(61!)}{{2}^{61}\cdot 31(31!)(30!)}$$=2(4⋅6⋅8⋯62)262!​=261⋅(31!)262!​=261⋅31(31!)(30!)62(61!)​

$$={2}^{-60}{\cdot }^{61}{C}_{30}$$=2−60⋅61C30​

$$={2}^{-60}\cdot C3161$$=2−60⋅C3161

Now on comparing with ${2}^{\alpha }(C3161)$2α(C3161)

We get $\alpha =-60$α=−60