Finding the coefficients of $x^0,x^1,x^2, \dots ,x^{10}$ in $(1+x^2+x^5)^{100}$

Question

I have been able to find $x_0: 1,$ $x_1: 0,$ $x_2: {100\choose 1} = 100,$ $x_3: 0,$ $x_4: {100\choose 2} = 4950$, and $x_5: {100 \choose 1 }= 100$. I am having trouble with figuring out how to get the rest up to $x_{10}$. Any suggestions?

Answer 1

The binomial series is

$$ (1+y)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} y^k. $$

With $y=x^2 + x^5$ and $\alpha=100$ you have $$ (1+x^2+x^5)^{100} = \sum_{l=0}^\infty {100 \choose l} x^{2l}(1 + x^3)^l. $$ Now with $y=x^3$ and $\alpha = l$ you have $$ (1+x^2+x^5)^{100} = \sum_{l=0}^\infty {100 \choose l} x^{2l} \sum_{k=0}^\infty {l \choose k} x^{3k}. $$ Simplifying, one gets $$ (1+x^2+x^5)^{100} = \sum_{l=0}^\infty \sum_{k=0}^\infty {100 \choose l}{l \choose k} x^{2l+3k}. $$ From here you can pick off any coefficient. For example, the coefficient of $x^{15}$ will come from terms with $15=2l+3k$, meaning $(l,k) \in \{(6,1),(0,5),(3,3)\}$, so this coefficient is $$ {100\choose 6}{6\choose1}+{100\choose 5}{0\choose5}+{100\choose3}{3\choose 3}$$