what is the co-efficient of $x^{50}$ in the expansion of
$$\frac{1}{(1-x^{1.7})(1-x^{1.8})(1-x^{2.6})(1-x^{3.0})(1-x^{4.0})(1-x^{6.7})(1-x^{7.5})(1-x^{8.2})}$$
can you please explain me the logic
what is the co-efficient of $x^{50}$ in the expansion of
$$\frac{1}{(1-x^{1.7})(1-x^{1.8})(1-x^{2.6})(1-x^{3.0})(1-x^{4.0})(1-x^{6.7})(1-x^{7.5})(1-x^{8.2})}$$
can you please explain me the logic
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Using the expansion $$\frac{1}{1-x}=1+x+x^2+x^3+\cdots\ ,$$ your product becomes $$(1+x^{1.7}+x^{2\times1.7}+x^{3\times1.7}+\cdots) (1+x^{1.8}+x^{2\times1.8}+x^{3\times1.8}+\cdots)\cdots\ .$$ In expanding this, each term in the answer comes from multiplying one term out of each bracket. For example, one of the terms is $$(x^{1.7})(x^{2\times1.8})(x^{2.6})(1)(x^{5\times3.0})(1)(1)(x^{7.5})(1) =x^{30.4}\ .$$ You need to find out how many ways there are of doing this to get $x^{50}$ as your final answer. I can't see any easy way of doing this, though a couple of fairly obvious possibilities are $$(1)(1)(1)(x^{2\times3.0})(x^{11\times4.0})(1)(1)(1)=x^{50}$$ and $$(1)(x^{5\times1.8})(1)(1)(1)(1)(1)(x^{5\times8.2})=x^{50}\ .$$