The value of the determinant `|{:(1,1,1,1),(1,2,3,4),(1,3,6,10),(1,4,10,20):}|` is

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{vmatrix} \]

we can use row operations to simplify the determinant.

### Step 1: Apply Row Operations
We will perform the following row operations:
- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - R_1 \)
- \( R_4 \leftarrow R_4 - R_1 \)

This gives us:

\[ D = \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 2 & 5 & 9 \\ 0 & 3 & 9 & 19 \end{vmatrix} \]

### Step 2: Expand the Determinant
Now we can expand the determinant along the first column:

\[ D = 1 \cdot \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 9 \\ 3 & 9 & 19 \end{vmatrix} \]

### Step 3: Calculate the 3x3 Determinant
Next, we calculate the determinant of the 3x3 matrix:

\[ \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 9 \\ 3 & 9 & 19 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix, we have:

\[ = 1 \cdot (5 \cdot 19 - 9 \cdot 9) - 2 \cdot (2 \cdot 19 - 9 \cdot 3) + 3 \cdot (2 \cdot 9 - 5 \cdot 3) \]

Calculating each term:

1. First term: \( 5 \cdot 19 - 9 \cdot 9 = 95 - 81 = 14 \)
2. Second term: \( 2 \cdot 19 - 9 \cdot 3 = 38 - 27 = 11 \) (multiplied by -2 gives \(-22\))
3. Third term: \( 2 \cdot 9 - 5 \cdot 3 = 18 - 15 = 3 \) (multiplied by 3 gives \(9\))

Putting it all together:

\[ = 1 \cdot 14 - 22 + 9 = 14 - 22 + 9 = 1 \]

### Final Result
Thus, the value of the determinant is

\[ D = 1 \]