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Question

(A) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is given by: $T_{r+1} = { }^{18} C_r (\sqrt{x})^{18

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$-{ }^{18} C_9 2^9$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is given by: $T_{r+1} = { }^{18} C_r (\sqrt{x})^{18-r} \left(-\frac{2}{\sqrt{x}}\right)^r$ $T_{r+1} = { }^{18} C_r (x)^{\frac{18-r}{2}} (-2)^r (x)^{-\frac{r}{2}}$ $T_{r+1} = { }^{18} C_r (-2)^r x^{\frac{18-r-r}{2}}$ $T_{r+1} = { }^{18} C_r (-2)^r x^{9-r}$ For the term to be independent of $x$,the exponent of $x$ must be $0$: $9 - r = 0 \Rightarrow r = 9$ Substituting $r = 9$ into the general term: $T_{9+1} = { }^{18} C_9 (-2)^9$ $T_{10} = -{ }^{18} C_9 2^9$

The term independent of $x$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is

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