Given an+2=an+12+an
⇒an+2an+1−anan+1=2
Now, Tr=ar+1ar is an A.P. with common difference 2
Now, T1=a1a2=2,T2=a2a3=4,....,Tr=2r
So (a3a1+a21)⋅(a4a2+a31)⋯(a32a30+a311)=i=1∏30(ai+2ai+ai+11)
=i=1∏30(ai+1ai+2aiai+1+1)=i=1∏30Tr+1Tr+1
=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
=2(4⋅6⋅8⋯62)262!=261⋅(31!)262!=261⋅31(31!)(30!)62(61!)
=2−60⋅61C30
=2−60⋅C3161
Now on comparing with 2α(C3161)
We get α=−60