In the expansion of left(1+frac{3x}{2}right)^ | vedclass

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(D) The general term in the expansion of $(1+x)^k$ is given by $\frac{k(k-1)...(k-r+1)}{r!} x^r$. For $\left(1+\frac{3x}{2}\right)^{-5}$,the coefficie

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$21$

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(D) The general term in the expansion of $(1+x)^k$ is given by $\frac{k(k-1)...(k-r+1)}{r!} x^r$. For $\left(1+\frac{3x}{2}ight)^{-5}$,the coefficient of $x^{10}$ is: $\frac{(-5)(-6)...(-5-10+1)}{10!} 	imes \left(\frac{3}{2}ight)^{10} = \frac{(-1)^{10} \cdot 5 \cdot 6 \cdot ... \cdot 14}{10!} 	imes \frac{3^{10}}{2^{10}} = \binom{14}{10} \left(\frac{3}{2}ight)^{10}$. In the expansion of $(1+ax)^n$,the coefficient of $x^{10}$ is $\binom{n}{10} a^{10}$. Equating the two: $\binom{14}{10} \left(\frac{3}{2}ight)^{10} = \binom{n}{10} a^{10}$. Comparing the terms,we get $n=14$ and $a=\frac{3}{2}$. However,checking the provided solution logic: $\binom{14}{10} = \binom{14}{4} = \frac{14 \cdot 13 \cdot 12 \cdot 11}{4 \cdot 3 \cdot 2 \cdot 1} = 1001$. Given the options and the standard interpretation of such problems,if $n=14$ and $a=1.5$,$na = 21$. Thus,$na = 21$.

In the expansion of $\left(1+\frac{3x}{2}\right)^{-5}$,the coefficient of $x^{10}$ is equal to the coefficient of $x^{10}$ in $(1+ax)^n$,where $n \in N$. Then $na$ is equal to:

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