If the $2 \mathrm{nd}, 3 \mathrm{rd}$ and 4 th terms in the expansion of $(x+a)^n$ are $96,216,216$ respectively and $n$ is a positive integer, then $a+x=$

If $|x|<1$, then the number of terms in the expansion of $\left[\frac{1}{2}\left(1 \cdot 2+2 \cdot 3 x+3 \cdot 4 x^2+\ldots . \infty\right)\right]^{-25}$

If the ratio of the terms equidistant from the middle term in the expansion of $(l+x)^{12}$ is $\frac{1}{256}(x \in N)$, then sum of all the terms of the expansion $(1+x)^{12}$ is