The coefficient of `t^4` in the expansion of `((1 - t^6)/(1-t))^(3)` is 3k. The value of k is _________.

Explanation

To find the coefficient of \( t^4 \) in the expansion of \( \left( \frac{1 - t^6}{1 - t} \right)^3 \), we can break the problem down into manageable steps.

### Step 1: Rewrite the expression
We start with the expression:
\[ \left( \frac{1 - t^6}{1 - t} \right)^3 = (1 - t^6)^3 (1 - t)^{-3} \]

### Step 2: Expand \( (1 - t^6)^3 \)
Using the binomial theorem, we can expand \( (1 - t^6)^3 \):
\[ (1 - t^6)^3 = \sum_{k=0}^{3} \binom{3}{k} (-t^6)^k = 1 - 3t^6 + 3t^{12} - t^{18} \]

### Step 3: Expand \( (1 - t)^{-3} \)
Using the generalized binomial series, we expand \( (1 - t)^{-3} \):
\[ (1 - t)^{-3} = \sum_{n=0}^{\infty} \binom{n + 2}{2} t^n \]
This gives us the coefficients for \( t^n \) as \( \binom{n + 2}{2} \).

### Step 4: Combine the expansions
Now we need to find the coefficient of \( t^4 \) in the product of the two expansions:
\[ (1 - 3t^6 + 3t^{12} - t^{18}) \cdot \left( \sum_{n=0}^{\infty} \binom{n + 2}{2} t^n \right) \]

### Step 5: Identify the relevant terms
To find the coefficient of \( t^4 \), we consider the contributions from the expansion:
- From \( 1 \): contributes \( \binom{4 + 2}{2} = \binom{6}{2} = 15 \)
- From \( -3t^6 \): contributes nothing since \( t^6 \) cannot combine with any term to yield \( t^4 \).
- From \( 3t^{12} \): contributes nothing since \( t^{12} \) cannot combine with any term to yield \( t^4 \).
- From \( -t^{18} \): contributes nothing since \( t^{18} \) cannot combine with any term to yield \( t^4 \).

### Step 6: Calculate the total coefficient
The only contribution to \( t^4 \) comes from the \( 1 \) term:
\[ \text{Coefficient of } t^4 = 15 \]

### Step 7: Relate to the given equation
According to the problem, the coefficient of \( t^4 \) is given as \( 3k \):
\[ 3k = 15 \]

### Step 8: Solve for \( k \)
Dividing both sides by 3:
\[ k = \frac{15}{3} = 5 \]

Thus, the value of \( k \) is \( \boxed{5} \).