The ratio of the coefficient of x^2 to the co | vedclass

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Question

(B) Given expression: $(x^5 + 4 \cdot 3^{-\log_{\sqrt{3}}\sqrt{x^3}})^{10}$Simplify the term: $3^{-\log_{\sqrt{3}}\sqrt{x^3}} = 3^{-\log_{3^{1/2}}x^{3

Vedclass · Accepted Answer

$10:3$

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(B) Given expression: $(x^5 + 4 \cdot 3^{-\log_{\sqrt{3}}\sqrt{x^3}})^{10}$Simplify the term: $3^{-\log_{\sqrt{3}}\sqrt{x^3}} = 3^{-\log_{3^{1/2}}x^{3/2}} = 3^{-2 \cdot \frac{3}{2} \log_3 x} = 3^{-3 \log_3 x} = x^{-3}$.So,the expression becomes $(x^5 + 4x^{-3})^{10}$.The general term $T_{r+1} = {}^{10}C_r (x^5)^{10-r} (4x^{-3})^r = {}^{10}C_r \cdot 4^r \cdot x^{50-5r-3r} = {}^{10}C_r \cdot 4^r \cdot x^{50-8r}$.For the coefficient of $x^2$: $50-8r = 2$ $\Rightarrow 8r = 48$ $\Rightarrow r = 6$.Coefficient of $x^2 = {}^{10}C_6 \cdot 4^6$.For the coefficient of $x^{10}$: $50-8r = 10$ $\Rightarrow 8r = 40$ $\Rightarrow r = 5$.Coefficient of $x^{10} = {}^{10}C_5 \cdot 4^5$.Ratio = $\frac{{}^{10}C_6 \cdot 4^6}{{}^{10}C_5 \cdot 4^5} = \frac{210 \cdot 4}{252} = \frac{840}{252} = \frac{10}{3}$.

The ratio of the coefficient of $x^2$ to the coefficient of $x^{10}$ in the expansion of $(x^5 + 4 \cdot 3^{-\log_{\sqrt{3}}\sqrt{x^3}})^{10}$ is

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