The square root of independent term in the expansion of $\left(\frac{2…
The square root of independent term in the expansion of $\left(\frac{2 x^2}{5}+\sqrt{\frac{5}{x}}\right)^{10}$ is
- $15 \sqrt{10}$
- $10 \sqrt{15}$
- $30 \sqrt{5}$
- $20 \sqrt{5}$
Solution
Since the general term in the expansion is $\left(\frac{2 x^2}{5}+\sqrt{\frac{5}{x}}\right)^{10}$
$\mathrm{T}_{r+1}={ }^{10} \mathrm{C}_r\left(\frac{2 x^2}{5}\right)^{10-r}\left(\frac{5^{\frac{1}{2}}}{x^{\frac{1}{2}}}\right)^r={ }^{10} \mathrm{C}_r \frac{2^{10-r} x^{20-2 r-\frac{r}{2}}}{5^{10-r-\frac{r}{2}}}$
for independent of $x, 20-2 r-\frac{r}{2}=0 \Rightarrow r=8$
So, $\mathrm{T}_9={ }^{10} \mathrm{C}_8 \frac{2^2}{5^{-2}}=\frac{10 \times 9}{2} \times 4 \times 5^2=5^3 \times 3^2 \times 2^2$
Now, $\sqrt{T_9}=\sqrt{5^3 \times 3^2 \times 2^2}=5 \times 3 \times 2 \sqrt{5}=30 \sqrt{5}$
Asked in: AP EAMCET 2024 (18 May Shift 1)
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