Using binomial thaorem (http://en.wikipedia.org/wiki/Binomial_theorem) find the general formula for the coefficients of the expantion: $$ \left(\sum_{i=0}^{\infty}\frac{t^{2i}}{n^i6^ii!}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)\right)^n $$
Thank you for your help.
The inner sum is $$ \sum_{i=0}^\infty\frac{x^{2i}}{n^i6^ii!}=e^{\frac{x^2}{6n}}\tag{1} $$ When raised to the $n^{\text{th}}$ power, $(1)$ is $e^{x^2/6}$. So it remains to compute $$ e^{x^2/6}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)^n =e^{x^2/6}\sum_{j=0}^n\binom{n}{j}\left(-\frac{t^2}{6}+\frac{t^4}{120}\right)^j\tag{2} $$ If the desire is to $(2)$ up to $O(t^{2k})$, one only need sum the first $k$ terms; that is, $$ e^{x^2/6}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)^n=e^{x^2/6}\sum_{j=0}^{k-1}\binom{n}{j}\left(-\frac{t^2}{6}+\frac{t^4}{120}\right)^j+e^{x^2/6}O(t^{2k})\tag{3} $$