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)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))` (b) `(n(n+1))/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))` (c) `(n(n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))` (d) None of these

Options

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

Answer

A