<s>[INST] The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:

In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
  coefficients of $x^3$?

 [/INST] The multinomial expansion gives $\mathbf{S}= (1 - \frac{x}{2} + \frac{1}{\sqrt{x}})^8 = \sum_{k_1+k_2 +k_3=8} \binom{8}{k_1,k_2,k_3} (1)^{k_1}(\frac{-x}{2})^{k_2} (\frac{1}{\sqrt{x}})^{k_3}.$ Simplifying we get $\mathbf{S}=\sum_{k_1+k_2 +k_3=8} \binom{8}{k_1,k_2,k_3} (\frac{-1}{2})^{k_2} (x)^{k_2 -k_3/2}.$ Now identify the tuples for which $(k_1,k_2,k_3) \in \mathbb{Z}_{+}^{3}$ such that $k_1+k_2+k_3=8$ and $k_2-k_3/2=3.$ Then add the coefficients for these tuples to get your answer.
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