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?
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3$\begingroup$ Have you tried raising your expression to the second, third, fourth power? If not I would suggest doing that and attempting to observe a pattern $\endgroup$– David PMar 26 at 0:36
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1$\begingroup$ Apart from $x^{10}$ there is at most one way of making each power up to order. $\endgroup$– Zoe AllenMar 26 at 0:48
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1$\begingroup$ In other words, six must be three twos; seven must be a five and a two; and so on. $\endgroup$– Gerry MyersonMar 26 at 0:51
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1 Answer
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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 $$ {100\choose 6}{6\choose1}+{100\choose 5}{0\choose5}+{100\choose3}{3\choose 3}$$
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$\begingroup$ There’s some mixing between l and k towards the end, e.g. is l = 0 or k = 0? $\endgroup$– yurneroMar 26 at 1:23
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$\begingroup$ I think it's correct noting ${0\choose 5}=0$ but please feel free to edit any mistakes ! $\endgroup$ Mar 26 at 1:35