The ratio of the coefficient of x^{15} to the | vedclass

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(D) The general term in the expansion of $(x^2 + \frac{2}{x})^{15}$ is given by $T_{r+1} = ^{15}C_r (x^2)^{15-r} (2x^{-1})^r = ^{15}C_r \cdot 2^r \cdo

Vedclass · Accepted Answer

$1: 32$

Vedclass · Answer

(D) The general term in the expansion of $(x^2 + \frac{2}{x})^{15}$ is given by $T_{r+1} = ^{15}C_r (x^2)^{15-r} (2x^{-1})^r = ^{15}C_r \cdot 2^r \cdot x^{30-3r}$.For the term independent of $x$,we set the exponent of $x$ to $0$:$30 - 3r = 0 \Rightarrow r = 10$.Thus,the independent term is $T_{11} = ^{15}C_{10} \cdot 2^{10}$.For the coefficient of $x^{15}$,we set the exponent of $x$ to $15$:$30 - 3r = 15$ $\Rightarrow 3r = 15$ $\Rightarrow r = 5$.Thus,the coefficient of $x^{15}$ is $^{15}C_5 \cdot 2^5$.The required ratio is $\frac{^{15}C_5 \cdot 2^5}{^{15}C_{10} \cdot 2^{10}}$.Since $^{15}C_5 = ^{15}C_{10}$,the ratio simplifies to $\frac{2^5}{2^{10}} = \frac{1}{2^5} = \frac{1}{32}$ or $1:32$.

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of $(x^2 + \frac{2}{x})^{15}$ is

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