The square root of the independent term in th | vedclass

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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{2x^2}{5} + \left(\frac{5}{x}\right)^{1/2}\right)^{10}$ is given by:$T_{r+1} = {}^{10}C

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$30\sqrt{5}$

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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{2x^2}{5} + \left(\frac{5}{x}ight)^{1/2}ight)^{10}$ is given by:$T_{r+1} = {}^{10}C_r \left(\frac{2x^2}{5}ight)^{10-r} \left(\frac{5^{1/2}}{x^{1/2}}ight)^r$$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r}} \cdot x^{2(10-r)} \cdot \frac{5^{r/2}}{x^{r/2}}$$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r-r/2}} \cdot x^{20-2r-r/2}$For the term to be independent of $x$,the exponent of $x$ must be zero:$20 - 2r - \frac{r}{2} = 0$ $\Rightarrow 20 = \frac{5r}{2}$ $\Rightarrow r = 8$Substituting $r=8$ into the expression:$T_9 = {}^{10}C_8 \cdot \frac{2^{10-8}}{5^{10-8-4}} = {}^{10}C_2 \cdot \frac{2^2}{5^{-2}} = 45 \cdot 4 \cdot 25 = 4500$The square root of the independent term is $\sqrt{4500} = \sqrt{900 	imes 5} = 30\sqrt{5}$.

The square root of the independent term in the expansion of $\left(\frac{2x^2}{5} + \sqrt{\frac{5}{x}}\right)^{10}$ is

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