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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.

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Seems like a tedious exercise, but why would you get stuck? What have you tried? –  TMM Jun 19 '12 at 22:41
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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, $$ e^{x^2/6}\left(1-\frac{t^2}{6}+\frac{t^4}{120}\right)^n=e^{x^2/6}\sum_{j=0}^{k-1}\binom{n}{j}\left(-\frac{t^2}{6}+\frac{t^4}{120}\right)^j+e^{x^2/6}O(t^{2k})\tag{3} $$

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I am so sorry, I've just noticed that everything should be of variable $t$. –  Aleks.K Jun 20 '12 at 0:38
    
Is it possible to fins a general formula for the coefficient of the expansion for $$ \left(\sum_{i=0}^{\infty}\frac{t^{2i}}{n^i6^ii!}\left(1-\frac{t^2}{6}+\frac{t^4}‌​{120}\right)\right)^n $$ by first multiplying $(1-t^2/6+t^4/120)$ by the series of $e^{t^2/(6n)}$ and then using binomial theorem? –  Aleks.K Jun 20 '12 at 0:41
    
@Aleks.K: It is possible to multiply two power series to get a new power series. This can be done repeatedly to get the $n^{\text{th}}$ power. If the constant term is not $0$ and you multiply two power series mod $t^n$, you get the product series mod $t^n$. If the lowest power term in the power series is $t^k$, then you need to use the power series mod $t^{n+k}$. –  robjohn Jun 20 '12 at 1:08
    
@Aleks.K: For example, $$(1+t+3t^2+O(t^3))(t^2+2t^3-t^4+O(t^5))=t^2+3t^3+4t^4+O(t^5)$$ –  robjohn Jun 20 '12 at 1:13
    
Thank you for your eplanations. –  Aleks.K Jun 23 '12 at 13:11
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