10) A student of Fibonacci invented a sequence of numbers defined in this way: the first and second numbers of the sequence are 0 and 1, respectively; each number of the sequence, from the third onward, is equal to the sum of all the numbers that precede it (excluding itself!). What is the fifteenth number of the sequence?
(A) 377,
(B) 2084,
(C) 2584,
(D) 3012,
(E) 4096.
10. The answer is $(\mathbf{E})$.

If we calculate by hand the first terms of the sequence, we find:

$$
0,1,1,2,4,8,16,32, \ldots
$$

That is, except for the first term, they are all powers of 2, and more precisely, from the third term onward, we have

$$
F_{n}=2^{n-3},
$$

where $F_{n}$ denotes the $n$-th term of the sequence. This equality can be proven as follows: for $n \geq 3$, we can write

$$
\begin{aligned}
& F_{n}=F_{n-1}+\cdots+F_{0}, \\
& F_{n+1}=F_{n}+F_{n-1}+\cdots+F_{0}=F_{n}+F_{n}=2 F_{n} .
\end{aligned}
$$

Thus, each term of the sequence (from the fourth onward) is double the previous one. Since the third term is $F_{3}=1=2^{0}$, it follows that $F_{n}=2^{n-3}$. In particular, $F_{15}=2^{12}=4096$.