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

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?

• 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 Mar 26 at 0:36
• Apart from $x^{10}$ there is at most one way of making each power up to order. Mar 26 at 0:48
• In other words, six must be three twos; seven must be a five and a two; and so on. Mar 26 at 0:51

$$(1+y)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} y^k.$$
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 $${100\choose 6}{6\choose1}+{100\choose 5}{0\choose5}+{100\choose3}{3\choose 3}$$
• I think it's correct noting ${0\choose 5}=0$ but please feel free to edit any mistakes ! Mar 26 at 1:35