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1+1/3x+(1xx4)/(3xx6)x^2+(1xx4xx7)/(3xx6x...

`1+1/3x+(1xx4)/(3xx6)x^2+(1xx4xx7)/(3xx6xx9)x^3+`.... is equal to

A

x

B

`(1+x)^(1//3)`

C

`(1-x)^(1//3)`

D

`(1-x)^(-1//3)`

Text Solution

Verified by Experts

The correct Answer is:
D
Let, `(1+y)^(n) = 1+1/3x+(1xx4)/(3xx6)x^(2)+(1xx4xx7)/(3xx6xx9)x^(3)+"…."`
`= 1+ny + (n(n-1))/(2!) y^(2) + "….."`
Comparing the terms, we get
`ny = 1/3 x, (n(n-1))/(2!) y^(2) = (1xx4)/(3xx6) x^(2)`
Solving, `n = - 1//3, y = -x` Hence, the given series.
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