6 Let the sequence $\left\{a_{n}\right\}$ be an arithmetic sequence, and the sequence $\left\{b_{n}\right\}$ satisfies: $b_{1}=a_{1}, b_{2}=a_{2}+$ $a_{3}, b_{3}=a_{4}+a_{5}+a_{6}, \cdots$, if $\lim _{n \rightarrow \infty} \frac{b_{n}}{n^{3}}=2$, then the common difference $d$ of the sequence $\left\{a_{n}\right\}$ is
(A) $\frac{1}{2}$
(B) 1
(C) 2
(D) 4
$6 \mathrm{D}$ hint:
$$
\begin{aligned}
b_{n} & =a \frac{n(n-1)}{2}+1+a \frac{n(n-1)}{2}+2 \\
& =\frac{n}{2}\left(a \frac{n(n-1)}{2}+1+a \frac{n(n-1)}{2}+n\right) \\
& =\frac{n}{2}\left(a_{1}+\frac{n(n-1)}{2} d+a_{1}+\left(\frac{n(n-1)}{2}+n\right.\right. \\
& =\frac{n}{2}\left(2 a_{1}-d+n^{2} d\right),
\end{aligned}
$$

Thus,
$$
\lim _{n \rightarrow \infty} \frac{b_{n}}{n^{3}}=\lim _{n \rightarrow \infty} \frac{1}{2}\left(\frac{2 a_{1}-d}{n^{2}}+d\right)=\frac{d}{2}=2,
$$

Solving for $d$ gives $d=4$.