$10 \cdot 57$ The sequence $1, (1+2), \left(1+2+2^{2}\right), \cdots, \left(1+2+2^{2}+\cdots+2^{n-1}\right)$, the sum of the first $n$ terms expressed in terms of $n$ is
(A) $2^{n}$.
(B) $2^{n}-n$.
(C) $2^{n+1}-n$.
(D) $2^{n+1}-n-2$.
(E) $n \cdot 2^{n}$.
(23rd American High School Mathematics Examination, 1972)
[Solution] It is known that the $k$-th term of the sequence equals $1+2+2^{2}+\cdots+2^{k-1}=2^{k}-1$, so the sum of the first $n$ terms can be written as
$$
\begin{aligned}
& \left(2^{1}-1\right)+\left(2^{2}-1\right)+\left(2^{3}-1\right)+\cdots+\left(2^{n}-1\right) \\
= & \left(2+2^{2}+2^{3}+\cdots+2^{n}\right)-\underbrace{(1+1+1+\cdots+1}_{n \text { terms }}) \\
= & \left(2^{n+1}-2\right)-n \\
= & 2^{n+1}-n-2 .
\end{aligned}
$$

Therefore, the answer is $(D)$.