MathematicsHard178×since 2002Q5337Consider the sequence a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldotsa1,a2,a3,… such that a1=1,a2=2a_{1}=1, a_{2}=2a1=1,a2=2 and an+2=2an+1+ana_{n+2}=\frac{2}{a_{n+1}}+a_{n}an+2=an+12+an for n=1,2,3,….\mathrm{n}=1,2,3, \ldots .n=1,2,3,…. If (a1+1a2a3)⋅(a2+1a3a4)⋅(a3+1a4a5)…(a30+1a31a32)=2α(61C31)\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)(a3a1+a21)⋅(a4a2+a31)⋅(a5a3+a41)…(a32a30+a311)=2α(61C31), then α\alphaα is equal to :A−-−30B−-−31C−-−60D−-−61Check answerSkip