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  • The term independent of in the expansion of <img alt="\left\{\frac{x+1}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{x-1}{x

The term independent of x in the expansion of \left\{\frac{x+1}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{x-1}{x-\sqrt{x}}\right\}^{15} is

Option: 1

{ }^{15} C_7


Option: 2

{ }^{15} C_9


Option: 3

{ }^{15} C_5


Option: 4

None of these


Answers (1)

best_answer

Now, \frac{x+1}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}

$$ =\frac{\left(x^{1 / 3}\right)^3+1^3}{\left(x^{1 / 3}\right)^2-x^{1 / 3}+1^2}=x^{1 / 3}+1 $$ \left( a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right)

and \frac{x-1}{x-\sqrt{x}}=\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}+1}{\sqrt{x}}=1+x^{-1 / 2}

Hence, \left\{\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-\sqrt{x}}\right\}^{15}=\left\{x^{1 / 3}+1-\left(1+x^{-1 / 2}\right)\right\}^{15}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{15}

General term

\begin{aligned} T_{r+1} & ={ }^{15} C_r\left(x^{1 / 3}\right)^{15-r}\left(-x^{-1 / 2}\right)^r \\ & ={ }^{15} C_r x^{(15-r) / 3}(-1)^r x^{-r / 2} \\ & \left.={ }^{15} C_r(-1)^r x^{(15-r) / 3-(r / 2)}\right) \end{aligned}

This will be independent of x if \frac{15-r}{3}-\frac{r}{2}=0  i.e. if 30-2r-3r=0  i.e. if r = 6

The term independent of x is T_{6+1} \text { i.e. } T_7={ }^{15} C_6(-1)^6 \times{ }^0={ }^{15} C_6={ }^{15} C_9 \text {. } 

 

Posted by

seema garhwal

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