6. Given $\left\{a_{n}\right\}$ is an arithmetic sequence, $n \in \mathbf{N}$, and the common difference is $d$,
$$
p_{n}(x)=a_{0} C_{n}^{0}(1-x)^{n}+a_{1} C_{n}^{1} x(1-x)^{n-1}+a_{2} C_{n}^{2} x^{2}(1-x)^{n-2}+\cdots+a_{n} C_{n}^{n} x^{n}
$$

Simplify $p_{n}(x)=$ $\qquad$
6. $a_{0}+n d x \quad$ According to the formula $C_{n+1}^{k}=C_{n}^{k}+C_{n}^{k-1}$, we get
$$
\begin{aligned}
p_{n}(x)= & a_{0} C_{n-1}^{0}(1-x)^{n}+a_{1}\left(C_{n-1}^{1}+C_{n-1}^{0}\right) x \cdot(1-x)^{n-1}+\cdots+a_{n-1}\left(C_{n-1}^{n-1}+C_{n-1}^{n-2}\right) x^{n-1}(1-x)+ \\
& a_{n} C_{n-1}^{n-1} x^{n} \\
= & {\left[a_{0} C_{n-1}^{0}(1-x)^{n}+a_{1} C_{n-1}^{1} x(1-x)^{n-1}+\cdots+a_{n-1} C_{n-1}^{n-1} x^{n-1}(1-x)\right]+\left[\left(a_{1} C_{n-1}^{0} x(1-x)^{n-1}\right.\right.} \\
& \left.+\cdots+a_{n} C_{n-1}^{n-1} x^{n}\right] \\
= & (1-x) p_{n-1}(x)+\left[\left(a_{0}+d\right) C_{n-1}^{0} x(1-x)^{n-1}+\left(a_{1}+d\right) C_{n-1}^{1} x^{2}(1-x)^{n-2}+\cdots+\left(a_{n-1}+\right.\right. \\
& \left.d) C_{n-1}^{n-1} x^{n}\right] \\
= & (1-x) p_{n-1}(x)+x p_{n-1}(x)+d x \\
= & d x+p_{n-1}(x)
\end{aligned}
$$

And $p_{1}(x)=a_{0}(1-x)+a_{1} x=a_{0}+d x$, by induction we know $p_{n}(x)=a_{0}+n d x$.