8. Expand the binomial $\left(\frac{\sqrt{x+1}}{2 \sqrt[4]{x}}\right)^{n}$ in descending powers of $x$. If the coefficients of the first two terms form an arithmetic sequence, then the number of terms in the expansion where the power of $x$ is an integer is $\qquad$ .
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8.【Analysis and Solution】It is not difficult to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$, respectively. Since these three numbers form an arithmetic sequence, we have $2-\frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get: $n=8$ and $n=1$ (discard).

When $n=8$, $T_{r+1}=\mathrm{C}_{8}^{r}\left(\frac{1}{2}\right)^{r} x^{\frac{(16-3 r)}{4}}$, where $r=0,1, \cdots, 8$. $r$ must satisfy $4 \mid(16-3 r)$, so $r$ can only be $0,4,8$.