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(C) The general term $T_{r+1}$ in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.Here,$a = \frac{1}{2}x^{1/3}$,$b = x^{-1/5}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.Here,$a = \frac{1}{2}x^{1/3}$,$b = x^{-1/5}$,and $n = 8$.$T_{r+1} = {}^8C_r (\frac{1}{2}x^{1/3})^{8-r} (x^{-1/5})^r = {}^8C_r (\frac{1}{2})^{8-r} x^{\frac{8-r}{3}} x^{-\frac{r}{5}} = {}^8C_r (\frac{1}{2})^{8-r} x^{\frac{8-r}{3} - \frac{r}{5}}$.For the term to be independent of $x$,the exponent of $x$ must be $0$:$\frac{8-r}{3} - \frac{r}{5} = 0$ $\Rightarrow 5(8-r) - 3r = 0$ $\Rightarrow 40 - 5r - 3r = 0$ $\Rightarrow 8r = 40$ $\Rightarrow r = 5$.Substituting $r = 5$ into the expression:$T_{5+1} = {}^8C_5 (\frac{1}{2})^{8-5} (1)^5 = {}^8C_5 (\frac{1}{2})^3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} 	imes \frac{1}{8} = 56 	imes \frac{1}{8} = 7$.

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

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