For n in N,in the expansion of left(sqrt[4]{x | vedclass

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(B) The given expression is $\left(x^{-\frac{3}{4}}+a x^{\frac{5}{4}}\right)^n$. The general term $T_{r+1}$ is given by ${}^n C_r \left(x^{-\frac{3}{4

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(B) The given expression is $\left(x^{-\frac{3}{4}}+a x^{\frac{5}{4}}\right)^n$. The general term $T_{r+1}$ is given by ${}^n C_r \left(x^{-\frac{3}{4}}\right)^{n-r} \left(a x^{\frac{5}{4}}\right)^r = {}^n C_r a^r x^{-\frac{3n}{4} + \frac{3r}{4} + \frac{5r}{4}} = {}^n C_r a^r x^{-\frac{3n}{4} + 2r}$. For the term to be independent of $x$,the exponent of $x$ must be zero: $-\frac{3n}{4} + 2r = 0$ $\Rightarrow 2r = \frac{3n}{4}$ $\Rightarrow r = \frac{3n}{8}$. Since $r$ must be an integer,$n$ must be a multiple of $8$. The sum of binomial coefficients is $(1+a)^n$. Given $200 If $n=8$,then $200 The term independent of $x$ is ${}^n C_r a^r = 448$. Substituting $n=8$ and $r=\frac{3(8)}{8}=3$,we get ${}^8 C_3 a^3 = 448$. $56 a^3 = 448$ $\Rightarrow a^3 = 8$ $\Rightarrow a = 2$. Checking the sum of coefficients: $(1+2)^8 = 3^8 = 6561$,which is not between $200$ and $400$. Re-evaluating the sum of binomial coefficients: The sum of binomial coefficients is $2^n$. Given $200 Thus,$n=8$ and $a=2$ satisfy the conditions.

For $n \in N$,in the expansion of $\left(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}\right)^n$,the sum of all binomial coefficients lies between $200$ and $400$ and the term independent of $x$ is $448$. Then the value of $a$ is

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