For $(1 + x + x^2)^6$, find the term which has $x^6$ in it. 
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.

I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.

When expanding the product $(x^0 + x^1 + x^2)^6$, each term looks like $x^{k_1} x^{k_2} \cdots x^{k_6} = x^{k_1+k_2+\dots+k_6}$ where all $k_i\in \{0,1,2\}$. Hence, the coefficient of $x^6$ in the expansion is the number of six-tuples $(k_1,k_2,\dots,k_6)$ such that $k_1+k_2+\dots+k_6=6$ and all $k_i\in\{0,1,2\}$.
The number of $2$s in such a partition can be $0$, $1$, $2$ or $3$ and for each there is exactly one possible partition up to permuation:
\begin{align}
6 &= 1+1+1+1+1+1 \qquad\text{1 permutation}, \\
  &= 1+1+1+1+2+0 \qquad\text{$\frac{6!}{4!\,1!\,1!}=30$ permutations}, \\
  &= 1+1+2+2+0+0 \qquad\text{$\frac{6!}{2!\,2!\,2!}=90$ permutations}, \\
  &= 2+2+2+0+0+0 \qquad\text{$\frac{6!}{3!\,3!}=20$ permutations.}
\end{align}
Hence, in total there are $141$ summands equal to $x^6$ in the expansion, so the term you are looking for is $141 x^6$.
