The term independent of x in the expansion of | vedclass

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Question

(b) $(10 - r)\left( {\frac{1}{2}} \right) + r( - 2) = 0 \Rightarrow 5 - \frac{r}{2} - 2r = 0 \Rightarrow r = 2$ 

So the term independent of $x

Vedclass · Accepted Answer

$5\over4$

Vedclass · Answer

(b) $(10 - r)\left( {\frac{1}{2}} ight) + r( - 2) = 0 \Rightarrow 5 - \frac{r}{2} - 2r = 0 \Rightarrow r = 2$ 

So the term independent of $x$

${ = ^{10}}{C_2} 	imes {\left( {\frac{1}{3}} ight)^4}{\left( {\frac{3}{2}} ight)^2} = \frac{{10 	imes 9}}{{2 	imes 1}} 	imes \frac{1}{{3 	imes 3 	imes 2 	imes 2}} = \frac{5}{4}$

The term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ will be

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