## Task A-4.2.

In one row, the numbers $1,2, \ldots, 2016$ are written in sequence. In each subsequent row, the sums of two adjacent numbers are written in sequence. For example, in the second row, the numbers $3,5, \ldots$, 4031 are written. In the last row, there is only one number. What is that number?
## First Solution.

We prove by mathematical induction on $n$ that the numbers in the $n$-th row form an arithmetic sequence with a difference of $2^{n-1}$.

For $n=2$, the claim holds, as we can see directly from the fact that the second row contains consecutive odd numbers.

Assume the claim holds for $n-1$. Let $A, B$, and $C$ be three consecutive terms in the $(n-1)$-th row, and $A+B, B+C$ be consecutive numbers in the $n$-th row. By the induction hypothesis, we have $B-A=C-B=2^{n-2}$, so $(B+C)-(A+B)=(C-B)+(B-A)=2^{n-2}+2^{n-2}=2^{n-1}$. This shows that the numbers in the $n$-th row form an arithmetic sequence with a difference of $2^{n-1}$.

Let $a_{n}$ be the first element in the $n$-th row. For all $n \in\{1,2, \ldots, 2016\}$, we have

$$
a_{n}=a_{n-1}+a_{n-1}+2^{n-2}=2 a_{n-1}+2^{n-2}
$$

Using this relation and $a_{1}=1$, we can easily prove by induction that $a_{n}=(n+1) 2^{n-2}$.

Thus, the desired number is $a_{2016}=2^{2014} \cdot 2017$.