5 Let $a_{n}=2^{n}, b_{n}=n(n=1,2,3, \cdots)$, and $A_{n}, B_{n}$ be the sums of the first $n$ terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively. Let $c_{n}=a_{n} B_{n}+b_{n} A_{n}-a_{n} b_{n}$. Then the sum of the first 10 terms of the sequence $\left\{c_{n}\right\}$ is ( ).
(A) $2^{10}+53$
(B) $2^{11}+53$
(C) $110 \times\left(2^{9}-1\right)$
(D) $110 \times\left(2^{10}-1\right)$
(5) $\mathrm{D}$ Hint: When $n \geqslant 2$,
$$
\begin{aligned}
c_{n}= & \left(A_{n}-A_{n-1}\right) B_{n}+\left(B_{n}-B_{n-1}\right) A_{n} \\
& -\left(A_{n}-A_{n-1}\right)\left(B_{n}-B_{n-1}\right),
\end{aligned}
$$

Therefore, $c_{n}=A_{n} B_{n}-A_{n-1} B_{n-1}$, which gives
$$
\begin{aligned}
c_{10}= & A_{10} B_{10}-A_{9} B_{9}, \\
c_{9}= & A_{9} B_{9}-A_{8} B_{8}, \\
& \cdots \cdots \\
c_{2}= & A_{2} B_{2}-A_{1} B_{1},
\end{aligned}
$$

Also, $c_{1}=a_{1} b_{1}=A_{1} B_{1}$, so the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$ is $A_{n} B_{n}$. Thus, the sum of the first 10 terms of the sequence $\left\{c_{n}\right\}$ is
$$
2\left(2^{10}-1\right) \frac{10 \times 11}{2}=110\left(2^{10}-1\right) \text {. }
$$

Therefore, the answer is D.