`a"@"b=(a+b)(ab+b)`
`a#b=(a-b)(ab-b)`
`aDeltab=(a"@"b)-(a#b)`
Find the value of `((1"@"(2Delta3))#4)`:

Options

A. 35500
B. 482420
C. 123400
D. none of these

Answer

To solve the problem step by step, we will follow the definitions provided for the operations "@" (at), "#" (hash), and "Δ" (delta).

### Step 1: Calculate \( 2 \Delta 3 \)

Using the definition of \( a \Delta b = a"@b - a#b \):

1. **Calculate \( 2"@3 \)**:
\[ 2"@3 = (2 + 3)(2 \cdot 3 + 3) = (5)(6 + 3) = 5 \cdot 9 = 45 \]

2. **Calculate \( 2#3 \)**:
\[ 2#3 = (2 - 3)(2 \cdot 3 - 3) = (-1)(6 - 3) = -1 \cdot 3 = -3 \]

3. **Now calculate \( 2 \Delta 3 \)**:
\[ 2 \Delta 3 = 2"@3 - 2#3 = 45 - (-3) = 45 + 3 = 48 \]

### Step 2: Calculate \( 1"@(2 \Delta 3) \)

Now we need to calculate \( 1"@48 \):

1. **Calculate \( 1"@48 \)**:
\[ 1"@48 = (1 + 48)(1 \cdot 48 + 48) = (49)(48 + 48) = 49 \cdot 96 = 4704 \]

### Step 3: Calculate \( (1"@(2 \Delta 3))#4 \)

Now we need to calculate \( 4704#4 \):

1. **Calculate \( 4704#4 \)**:
\[ 4704#4 = (4704 - 4)(4704 \cdot 4 - 4) = (4700)(18816 - 4) = 4700 \cdot 18812 \]

2. **Now calculate \( 4700 \cdot 18812 \)**:
\[ 4700 \cdot 18812 = 88408400 \]

### Final Answer

Thus, the value of \( ((1"@(2 \Delta 3))#4) \) is \( 88408400 \).