5. Given the sequence $\left\{a_{n}\right\}$ is 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
5. D.

Notice,
$$
\begin{array}{l}
b_{n}=a \frac{n(n-1)}{2}+1+a \frac{n(n-1)}{2}+2+\cdots+a \frac{n(n-1)}{2}+n \\
=\frac{n}{2}\left[a_{\frac{n(n-1)}{2}}^{2}+1+a_{\frac{n(n-1)}{2}}^{2}+n\right. \\
=\frac{n}{2}\left\{a_{1}+\frac{n(n-1)}{2} d+a_{1}+\left[\frac{n(n-1)}{2}+n-1\right] d\right\} \\
=\frac{n}{2}\left(2 a_{1}-d+n^{2} d\right) .
\end{array}
$$

Then $\lim _{n \rightarrow \infty} \frac{b_{n}}{n^{3}}=\lim _{n \rightarrow \infty} \frac{1}{2}\left(\frac{2 a_{1}-d}{n^{2}}+d\right)=\frac{d}{2}=2$.
Thus, $d=4$.