<s>[INST] $n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
 [/INST] You should be able to do the third term in your decomposition.
For the second term, follow Mark Bennet's hint:  write the fraction as $$\frac{x+2}{9(1+x+x^2)} = \frac{(x+2)(1-x)}{9(1-x^3)} = \frac{2-x-x^2}{9} \cdot \frac{1}{1-x^3},$$ then write the second factor as an infinite geometric series with common ratio $x^3$.
For the first term, you need to observe that $$f(x) = \frac{1}{1-x} = \sum_{k=0}^\infty x^k$$ implies $$f'(x) = \frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1},$$ and $$f''(x) = \frac{2}{(1-x)^3} = \sum_{k=2}^\infty k(k-1) x^{k-2}.$$  Then you have to put this all together and collect like powers in $x$.  Tedious, but computationally feasible.
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