<s>[INST] Coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$ 
What is the coefficient of coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$?

Using summation of G.P., this is equivalent to finding the coefficient of $x^{41}$ in
$$\left(x^5 \left(\frac{1-x^5}{1-x}\right)\right)^5$$
and thus finding coefficient of $x^{16}$ in $(\frac{1-x^5}{1-x})^5$. How to proceed after this?
 [/INST] Why not to apply the multinomial sum (just to study it a little bit) so we need the coefficient of $x^{16}$ in $(x^4+x^3+x^2+x+1)^5$?
Put $A=x^3+x^2+x+1$ so one has $$(x^4+A)^5=x^{20}+5x^{16}A+10x^{12}A^2+10x^8A^3+5x^4A^4+A^5$$
Hence, after to see in $A,A^2,A^3,A^4$ ($A^5$ does not participate here because its greatest exponent is $15$) the coefficients of $1,x^4,x^8,x^{12}$ respectively, the required coefficient is equal to
$$\color{red}{5\cdot1+10\cdot3+10\cdot3+5\cdot1=70}$$
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