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If `a_1,a_2,a_3,...` are in A.P. and `a_i>0` for each i, then `sum_(i=1)^n n/(a_(i+1)^(2/3)+a_(i+1)^(1/3)a_i^(1/3)+a_i^(2/3))` is equal to

A

`(n)/(a_(n+1)^(2//3)+a_(1)^(2//3)+a_(n+1)^(1//3)a_(1)^(1//3))`

B

`(n+1)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))`

C

`(n^(2))/(a_(n+1)^(2//3)+a_(1)^(2//3)+a_(n+1)^(1//3)a_(1)^(1//3))`

D

none of these

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The correct Answer is:
A
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