The term independent of x in the expansion of | vedclass

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(c) $\frac{1}{3}(8 - r) + r\left( { - \frac{1}{5}} \right) = 0 \Rightarrow \frac{8}{3} - \frac{r}{3} - \frac{r}{5} = 0 \Rightarrow r = 5$

Thus term

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(c) $\frac{1}{3}(8 - r) + r\left( { - \frac{1}{5}} ight) = 0 \Rightarrow \frac{8}{3} - \frac{r}{3} - \frac{r}{5} = 0 \Rightarrow r = 5$

Thus term independent of $x = {\,^8}{C_5} 	imes {\left( {\frac{1}{2}} ight)^3}{(1)^5}$

$ = \frac{{8 	imes 7 	imes 6}}{{3 	imes 2 	imes 1}} 	imes \frac{1}{8} = 7$

The term independent of $x$ in the expansion of ${\left( {\frac{1}{2}{x^{1/3}} + {x^{ - 1/5}}} \right)^8}$ will be

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