When |x|3,the coefficient of frac{1}{x^n} in | vedclass

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(B) Given expression is $x^{3/2}(3+x)^{1/2}$.Since $|x|>3$,we can write this as $x^{3/2} \cdot x^{1/2} (1 + \frac{3}{x})^{1/2} = x^2 (1 + \frac{3}{x})

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$(-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2^{n+2}(n+2)!} 3^{n+2}$

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(B) Given expression is $x^{3/2}(3+x)^{1/2}$.Since $|x|>3$,we can write this as $x^{3/2} \cdot x^{1/2} (1 + \frac{3}{x})^{1/2} = x^2 (1 + \frac{3}{x})^{1/2}$.Using the binomial expansion $(1+z)^k = \sum_{r=0}^{\infty} \binom{k}{r} z^r$,where $\binom{k}{r} = \frac{k(k-1)\dots(k-r+1)}{r!}$.Here $k = 1/2$ and $z = 3/x$.So,$x^2 (1 + \frac{3}{x})^{1/2} = x^2 \sum_{r=0}^{\infty} \binom{1/2}{r} (\frac{3}{x})^r = \sum_{r=0}^{\infty} \binom{1/2}{r} 3^r x^{2-r}$.We want the coefficient of $\frac{1}{x^n}$,so we set $2-r = -n$,which gives $r = n+2$.The coefficient is $\binom{1/2}{n+2} 3^{n+2}$.Calculating $\binom{1/2}{n+2} = \frac{(1/2)(1/2-1)\dots(1/2-(n+2)+1)}{(n+2)!} = \frac{(1/2)(-1/2)(-3/2)\dots(-(2n+1)/2)}{(n+2)!}$.$= \frac{(-1)^{n+1} \cdot 1 \cdot 3 \cdot 5 \dots (2n+1)}{2^{n+2} (n+2)!}$.Multiplying by $3^{n+2}$,the coefficient is $(-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2^{n+2}(n+2)!} 3^{n+2}$.

When $|x|>3$,the coefficient of $\frac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is

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