Answer to Question #301939 in Discrete Mathematics for Tege

Question

10. Find the coefficient of x²⁰ in A(x)=(x¹⁰+x¹¹+.....+x²⁵)(x¹+x²+....+x¹⁵)⁵(x²⁰+x²¹+.....+x⁴⁵)

Answer 1

Given, $A(x)=(x^{10}+x^{11}+\cdots+x^{25})(x^1+x^2+\cdots+x^{15})^{5}(x^{20}+x^{21}+\cdots+x^{45})$

$\begin{aligned} \text{Coefficient of $x^{20}$~in}~A(x) &= \text{Coefficient of $x^{20}$~in~} (x^{10}+x^{11}+\cdots+x^{25})\\ &\qquad(x^1+x^2+\cdots+x^{15})^{5}(x^{20}+x^{21}+\cdots+x^{45})\\ &= \text{Coefficient of $x^{20}$~in~} [x^{10} (1+x+\cdots+x^{15})\cdot \\&\qquad x^{5}(1+x+\cdots+x^{14})^{5}\cdot x^{20}(1+x+\cdots+x^{25})]\\ &=\text{Coefficient of $x^{20}$~in~} x^{35} [(1+x+\cdots+x^{15})\cdot \\&\qquad (1+x+\cdots+x^{14})^{5}\cdot(1+x+\cdots+x^{25})]\\ \end{aligned}$

The first term in the expansion of$x^{35} [(1+x+\cdots+x^{15})\cdot (1+x+\cdots+x^{14})^{5}\cdot(1+x+\cdots+x^{25})]$ will be $x^{35}$ and hence coefficient of $x^{20}$ in A(x) = 0, since negative powers cannot exist in the expansion of $(1+x+\cdots+x^{15})\cdot (1+x+\cdots+x^{14})^{5}\cdot(1+x+\cdots+x^{25}).$