11. The first two terms of a sequence are 1 and 2 . Each of the following terms in the sequence is the sum of all the terms which come before it in the sequence.
Which of these is not a term in the sequence?
A 6
B 24
C 48
D 72
E 96
Solution
D The third term of the sequence is $1+2=3$, the fourth term is $1+2+3=6$, the fifth term is $1+2+3+6=12$, and so on. So the sequence begins
$$
1,2,3,6,12, \ldots
$$

In fact it is not difficult to see that, from the fourth term onwards, each term is double the previous term. We show this as follows. Suppose $a$ is a term of the sequence other than the first two, and that $b$ is the term that immediately follows $a$. Then $a$ is the sum of all the terms that come before it. The term $b$ is also the sum of all the terms that become before it. Hence $b$ is the sum of all the terms that come before $a$, and $a$. So, $b=a+a=2 a$. Therefore each term from the fourth term onwards is double the previous term.

Since from the fourth term onwards each term is double the previous term, the first eight terms are
$$
1,2,3,6,12,24,48,96 \text {. }
$$

Since the terms continue to increase we see that of the given options, $6,24,48$ and 96 occur in the sequence, but 72 does not occur.

NOTE
The proof that from the fourth term onwards each term is twice the previous one can be given more formally, as follows.

We use the notation $x_{n}$ for the $n$th term of the sequence. We can then define the sequence by $x_{1}=1, x_{2}=2$, and for $n>2$,
$$
x_{n+1}=x_{1}+x_{2}+\cdots+x_{n} .
$$

It follows that, for $n>2$,
$$
\begin{aligned}
x_{n+2} & =x_{1}+x_{2}+\cdots+x_{n}+x_{n+1} \\
& =\left(x_{1}+x_{2}+\cdots+x_{n}\right)+x_{n+1} \\
& =x_{n+1}+x_{n+1} \\
& =2 x_{n+1} .
\end{aligned}
$$

This shows that from the fourth term onwards each term is double the previous term.