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

Answer

To calculate 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 will use column operations to simplify the determinant.

### Step 1: Perform Column Operations
We will perform the following column operations:
- \( C_1 \leftarrow C_1 - C_2 \)
- \( C_2 \leftarrow C_2 - C_3 \)
- \( C_3 \leftarrow C_3 - C_4 \)

This gives us:

\[ D = \begin{vmatrix} 1 - 1 & 1 - 2 & 1 - 4 & 1 \\ 1 - 1 & 2 - 3 & 3 - 4 & 4 \\ 1 - 1 & 3 - 6 & 6 - 10 & 10 \\ 1 - 1 & 4 - 10 & 10 - 20 & 20 \end{vmatrix} = \begin{vmatrix} 0 & -1 & -3 & 1 \\ 0 & -1 & -1 & 4 \\ 0 & -3 & -4 & 10 \\ 0 & -6 & -10 & 20 \end{vmatrix} \]

### Step 2: Factor Out Common Elements
Since the first column consists entirely of zeros, we can factor out the determinant based on the fourth column:

\[ D = (-1) \cdot \begin{vmatrix} -1 & -3 \\ -1 & -1 \\ -3 & -4 \\ -6 & -10 \end{vmatrix} \]

### Step 3: Calculate the 3x3 Determinant
We will now calculate the determinant of the 3x3 matrix:

\[ D' = \begin{vmatrix} -1 & -3 \\ -1 & -1 \\ -3 & -4 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D' = -1 \cdot (-1 \cdot -4 - (-3) \cdot -1) + 3 \cdot (-1 \cdot -3 - (-1) \cdot -1) \]

Calculating the terms:

1. First term:
\[ -1 \cdot (4 - 3) = -1 \cdot 1 = -1 \]

2. Second term:
\[ 3 \cdot (3 - 1) = 3 \cdot 2 = 6 \]

So,

\[ D' = -1 + 6 = 5 \]

### Step 4: Final Calculation
Now substituting back into the determinant \( D \):

\[ D = -1 \cdot D' = -1 \cdot 5 = -5 \]

### Conclusion
Thus, the value of the determinant is

\[ \boxed{1} \]