To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6$ in $(x+(1+i))^{10}(x+(1-i))^{10}$, then dividing by $2^{10}$.
Then, we find that the coefficient of $x^6$ would be:
$$\sum_{i=0}^{6} \binom{10}{6-i} \binom{10}{i} (1-i)^{10-(6-i)} (1+i)^{10-i}$$
with the knowledge that $(1-i)(1+i)=2$, I simplified to
$$\binom{10}{6}\binom{10}{0}2^4((1-i)^6+(1+i)^6)+\binom{10}{5}\binom{10}{1}2^5((1-i)^4+(1+i)^4)+\binom{10}{4}\binom{10}{2}2^6((1-i)^2+(1+i)^2)+\binom{10}{3}\binom{10}{3}2^7$$
Note: the formula $(1+i)^x+(1-i)^x$ gives:
$ 2(2^{\frac{x}{2}}) \cos(\frac{x\pi}{4})$
After simplifying and reapplying the division by $2^{10}$, I get $(\frac{0}{1024}) + (-8)(2520)(\frac{32}{1024}) + (\frac{0}{1024}) + (120)(120)(2^7)$, which gives $0-630+0+1800,$ which is 1170, and I checked this over with an expression expansion calculator.
If the original equation was $(1+x+x^2)^{10}$, I would have used binomials to find the answer, however, the $x^2$ was replaced by $\frac{x^2}{2}$.
My question is whether anyone has a combinatorics solution to this question, rather than just algebra. It would be nice if the solution did not require complex numbers.
Best Answer
The number of ways to partition 6 with just 2, 1, and 0 are \begin{align*} 6 &= (3, 0, 7)\cdot(2, 1, 0)\\ &= (2, 2, 6)\cdot(2, 1, 0)\\ &= (1, 4, 5)\cdot(2, 1, 0)\\ &= (0, 6, 4)\cdot(2, 1, 0) \end{align*} where $\cdot$ indicates the dot product. These partitions represent the choices of $x^2, x^1, x^0$ in the expansion of $(1 + x + x^2)^{10}$ to obtain $x^6$. However, we must account for the $x^2/2$, and so the sums are weighted by $(1/2)^n$, where $n$ is the number of times $x^2$ is chosen. Therefore, we have \begin{align*} \left(\frac{1}{2}\right)^3\binom{10}{3, 0, 7} + \left(\frac{1}{2}\right)^2\binom{10}{2, 2, 6} + \left(\frac{1}{2}\right)^1\binom{10}{1, 4, 5} + \left(\frac{1}{2}\right)^0\binom{10}{0, 6, 4} = 1170 \end{align*}