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

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Question

$T_{r+1}=^{15} C_{r}\left(x^{2}ight)^{15-r} \cdot\left(2 x^{-1}ight)^r$

$=15 \mathrm{C}_{r} 	imes(2)^{r} 	imes x^{30-3r}$

For independent

Vedclass · Accepted Answer

$1: 32$

Vedclass · Answer

$T_{r+1}=^{15} C_{r}\left(x^{2}ight)^{15-r} \cdot\left(2 x^{-1}ight)^r$

$=15 \mathrm{C}_{r} 	imes(2)^{r} 	imes x^{30-3r}$

For independent term, $30-3 r=0$

$\Rightarrow r=10$

Hence the term independent of $x$.

$\mathrm{T}_{11}=^{15} \mathrm{C}_{10} 	imes(2)^{10}$

For term involve $x^{15}, 30-3 r=15$

$ \Rightarrow r=5$

Hence coefficient of $x^{15}=^{15} C_{5} 	imes(2)^{5}$

Required ratio

$ = \frac{{{	ext{15}}{{	ext{C}}_5} 	imes {{(2)}^5}}}{{15{C_{10}} 	imes {{(2)}^{10}}}} = \frac{{\frac{{15!}}{{10!5!}}}}{{\frac{{15!}}{{5!10!}} 	imes {{(2)}^5}}}$

$ = 1:32$

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

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