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The pattern in the table can be summarized by the formula for the cell at row (i) and column (j) (where rows and columns start from 0):

(f(i,j) = (i+1)(j+1) - 1)

Explanation:

• For any cell ((i,j)):
  • The product ((i+1)(j+1)) represents the product of (row index +1) and (column index +1).
  • Subtract 1 to get the value in the cell.

Verification:

• Cell ((0,0)): ((0+1)(0+1)-1 =1-1=0) (matches).
• Cell ((1,1)): ((1+1)(1+1)-1=4-1=3) (matches).
• Cell ((2,3)): ((2+1)(3+1)-1=12-1=11) (matches).
• Cell ((5,5)): ((5+1)(5+1)-1=36-1=35) (matches).

Answer: The value at row (i) and column (j) is ((i+1)(j+1)-1).

If we denote rows and columns starting from 1 (let (r=i+1), (c=j+1)), the formula simplifies to (r \times c -1).

(\boxed{(i+1)(j+1)-1}) (or (\boxed{rc -1}) if using 1-based indices for rows (r) and columns (c)).

Assuming the problem expects the 1-based version (common in such tables), the answer is (\boxed{rc -1}). But let's confirm with the given table: for row 2 (1-based r=2) and column3 (1-based c=3), 2*3-1=5 which is exactly cell (1,2) in 0-based (row 1, column2: 5). Yes! So the 1-based formula is (\boxed{rc -1}).

Final Answer: (\boxed{rc -1}) (where (r) is row number and (c) is column number, starting from 1).

Alternatively, if 0-based is needed: (\boxed{(i+1)(j+1)-1}), but (\boxed{rc -1}) is more concise and likely expected.

(\boxed{rc -1})

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