Problem (Algebra):
The Sieve of Sundaram uses the following infinite table of positive integers:

| 4 | 7 | 10 | 13 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 12 | 17 | 22 | $\cdots$ |
| 10 | 17 | 24 | 31 | $\cdots$ |
| 13 | 22 | 31 | 40 | $\cdots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |

The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown.
Determine a formula for the number in the $R$ th row and $C$ th column.

Solution:
First, we determine the first entry in the $R$ th row.

Since the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.

Second, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.

Since the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.

Therefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is

$$
3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C
$$

Final answer: $2RC+R+C$