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

Monjur Alahi

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+1)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))`
C. `(n-1)/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))`
D. None of these


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Salim Ahmed Kawsar

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Correct Answer - C
`(c )` Let `d` be common difference of `A.P.impliesd=a_(i)-a_(i-1)`
Now `(1)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(a_(i+1)^(1//3)-a_(i)^(1//3))/(a_(i+1)-a_(i))`
`=(1)/(d)[a_(i+1)^(1//3)-a_(i)^(1//3)]`
Thus `sum_(i=1)^(n)(n)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(1)/(d)sum_(i=1)^(n-1)(a_(i+1)^(1//3)-a_(i)^(1//3))`
`=(1)/(d)(a_(n)^(1//3)-a_(1)^(1//3))`
`=(1)/(d)((a_(n)-a_(1)))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))`
`=((n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))`

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