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The coefficient of the term independent of `x` in the exampansion of `((x+1)/(x^(2//3)-x^(1//3)+1)-(x-1)/(x-x^(1//2)))^(10)` is `210` b. `105` c. `70` d. `112`

A

210

B

105

C

70

D

112

Text Solution

Verified by Experts

The correct Answer is:
a

We have,
= `(x+1) /(x^(2//3) - x^(1//3) + 1 )- (x -1)/(x - x ^(1//2))` ltbrge `((x^(1//3))^(3) + 1^(3)) /(x^(2//3) - x^(1//3) + 1 )- (x -1)/( x ^(1//2)-1)`
= `((x^(1//3)+ 1) x^(2//3) - x^(1//3) + 1 )/(x^(2//3) - x^(1//3) + 1 )- (x^(1//2) +1)/( x ^(1//2))`
` x^(1//3) + 1 + 1 - x ^(-1//2) = x^(1//3) - x^(-1//2)`
`((x+1)/(x^(2//3) - x^(1//3) + 1 )-(x-1)/ (x-x^(1//2)))^(10) = (x^(1//3) - x ^(-1//2))^(10)`
Let `T_(r +1)` be the general term in `(x^(1//3) - x ^(-1//2))^(10)`. Then,
`t_(r +1) = ""^(10)C_(r) (x^(1//3)^(10-r)(-1)^(r) (x ^(-1//2))^(r)`
For this term sto be independent of x, we must have
`(10 - r)/(3) - (r)/(2) = 0 rArr 20 - 2r - 3r = 0 rArr r = 4`
So, required coefficient = `""^(10)C_(4) (-1)^(4) = 210`
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OBJECTIVE RD SHARMA ENGLISH-BINOMIAL THEOREM AND ITS APPLCIATIONS -Section I - Solved Mcqs
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  19. Find the coefficient of x^5 in the expansion of (1+x)^(21)+(1+x)^(22)+...

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