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Find the co-efficient of $x^{18}$ in the expansion of $$(x+1)(x+2)...(x+10)(2x+1)(2x+3)...(2x+19)$$

What I've done : $$ (x+1)(x+2)...(x+10)(2x+1)(2x+3)...(2x+19) \\ =\frac{(2x+1)(2x+2)(2x+3)...(2x+20)}{2^{10}} $$ I can't think of any way to find the coefficient of $x^{18}$ in this expansion. What am I doing wrong ?

Vishnu V.S
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1 Answers1

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Now to get a term with $x^{18}$ you have to choose $18$ of the $2x$ terms and two of the constant terms, so your term in the numerator is $$(2x)^{18}\sum_{i=1}^{19}\sum_{j=i+1}^{20} ij$$

Ross Millikan
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