Find the co-efficient of x in the expansion of the determinant
`|{:((1+x)^2,(1+x)^4,(1+x)^6),((1+x)^3,(1+x)^6,(1+x)^9),((1+x)^4,(1+x)^8,(1+x)^12):}|`

Answer

To find the coefficient of \( x \) in the expansion of the determinant

\[ D = \begin{vmatrix} (1+x)^2 & (1+x)^4 & (1+x)^6 \\ (1+x)^3 & (1+x)^6 & (1+x)^9 \\ (1+x)^4 & (1+x)^8 & (1+x)^{12} \end{vmatrix} \]

we will follow these steps:

### Step 1: Write the determinant
We start by rewriting the determinant clearly:

\[ D = \begin{vmatrix} (1+x)^2 & (1+x)^4 & (1+x)^6 \\ (1+x)^3 & (1+x)^6 & (1+x)^9 \\ (1+x)^4 & (1+x)^8 & (1+x)^{12} \end{vmatrix} \]

### Step 2: Expand the determinant
We will use the properties of determinants and the binomial theorem to expand this determinant. The expansion of a determinant can be done by selecting one element from each row and multiplying them, while also accounting for the signs based on the position of the elements.

### Step 3: Calculate the contributions
Each term in the determinant will contribute a power of \( (1+x) \) based on the selected elements. We can express the determinant in terms of powers of \( (1+x) \):

1. The first term from the first row is \( (1+x)^2 \).
2. The second term from the second row is \( (1+x)^6 \) (from the second column).
3. The third term from the third row is \( (1+x)^{12} \) (from the third column).

The contributions from the other combinations will involve negative signs and will lead to different powers of \( (1+x) \).

### Step 4: Collect the powers of \( x \)
Using the binomial theorem, we know that:

\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

We are interested in the coefficient of \( x \), which corresponds to \( k=1 \) in the expansion.

### Step 5: Write down the coefficients
From the expansion, we can identify the coefficients of \( x \) from each term:

- From \( (1+x)^2 \): Coefficient of \( x \) is \( \binom{2}{1} = 2 \).
- From \( (1+x)^4 \): Coefficient of \( x \) is \( \binom{4}{1} = 4 \).
- From \( (1+x)^6 \): Coefficient of \( x \) is \( \binom{6}{1} = 6 \).
- From \( (1+x)^3 \): Coefficient of \( x \) is \( \binom{3}{1} = 3 \).
- From \( (1+x)^9 \): Coefficient of \( x \) is \( \binom{9}{1} = 9 \).
- From \( (1+x)^{12} \): Coefficient of \( x \) is \( \binom{12}{1} = 12 \).

### Step 6: Combine the coefficients
Now we will combine these coefficients based on the signs from the determinant expansion:

\[ \text{Coefficient of } x = 2 - 4 - 6 + 3 + 9 - 12 \]

### Step 7: Simplify the expression
Calculating the above expression step-by-step:

1. \( 2 - 4 = -2 \)
2. \( -2 - 6 = -8 \)
3. \( -8 + 3 = -5 \)
4. \( -5 + 9 = 4 \)
5. \( 4 - 12 = -8 \)

Thus, the coefficient of \( x \) in the expansion of the determinant is \( -8 \).

### Final Answer:
The coefficient of \( x \) is \( -8 \).

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