I'm trying to find to coefficient of $x^{100}$ of $$\sum_{n=0}^{∞}a_n x^{n}\ =\frac{1}{(1-x)(1-2x)(1-3x)}.$$ I used the sum: $$\frac{1}{1-x}\ = 1+x+x^2+\ldots.$$ So : $$\frac{1}{(1-x)(1-2x)(1-3x)}= \left(1+x+x^2+\ldots\right)\left(1+2x+(2x)^2+\ldots\right)\left(1+3x+(3x)^2+\ldots\right). $$ But multiplying out the right hand side to extract the coefficient $x^{100}$ is tedious. Any idea how I can obtain the coefficient in a simpler way ?
Regards
Use partial fractions: $$\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{1}{2} \cdot \frac{1}{1-x} - 4 \cdot \frac{1}{1-2x} + \frac{9}{2} \cdot \frac{1}{1-3x}.$$