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`1+1/3x+(1xx4)/(3xx6)x^2+(1xx4xx7)/(3xx6xx9)x^3+`is equal to`x`b. `(1+x)^(1//3)`c. `(1-x)^(1//3)`d. `(1-x)^(-1//3)`A. xB. `(1+x)^(1//3)`C. `(1-x)^(1//3)`D. `(1-x)^(-1//3)` |
Answer» Correct Answer - 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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