How do you find the coefficient of $$x^{12}$$ in $$(1-x^6)^4*(1-x)^{-4}$$
I mean this is just $$[1-{4\choose 1}x^6+{4\choose2}x^{12}-{4\choose3}x^{18}+{4\choose4}x^{24}]*[{-4\choose0} + {-4\choose1}(-x) + ... +{-4\choose12}(-x)^{12}]$$ How do I actually find that in there? Do I sum up all the terms up to it?
So, the coefficient of $x^{12}$
$=$ the coefficient of $x^0$ in $(1-x^6)^4 \cdot$ the coefficient of $x^{12}$ in $(1-x)^{-4}$
+the coefficient of $x^6$ in $(1-x^6)^4 \cdot$ the coefficient of $x^{12-6}$ in $(1-x)^{-4}$
+the coefficient of $x^{12}$ in $(1-x^6)^4 \cdot$ the coefficient of $x^{12-12}$ in $(1-x)^{-4}$
$$=\binom40\binom{-4}{12}-\binom41\binom{-4}6+\binom42\binom{-4}0$$
For sure, what lab bhattacharjee gave is the answer.
However, before starting, you could have simplified your problem since $$\frac{\left(1-x^6\right)^4}{(1-x)^4} = \left(x^5+x^4+x^3+x^2+x+1\right)^4$$
and the problem becomes simpler.
