If `(1+x+x^2+x^3)^n=sum_(r=0)^(3n)b_r x^r and sum_(r=0)^(3n)b_r=k ,"t h e n"sum_(r=0)^(3n)r b_r` is

Answer

To solve the problem, we need to find the value of \( \sum_{r=0}^{3n} r b_r \) given that \( (1+x+x^2+x^3)^n = \sum_{r=0}^{3n} b_r x^r \) and \( \sum_{r=0}^{3n} b_r = k \).

### Step-by-Step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ (1 + x + x^2 + x^3)^n \]
This can be rewritten as:
\[ \sum_{r=0}^{3n} b_r x^r \]

2. **Evaluate \( \sum_{r=0}^{3n} b_r \)**:
To find \( \sum_{r=0}^{3n} b_r \), we substitute \( x = 1 \):
\[ (1 + 1 + 1 + 1)^n = 4^n \]
Therefore, we have:
\[ \sum_{r=0}^{3n} b_r = k = 4^n \]

3. **Differentiate the Expression**:
Next, we differentiate \( (1 + x + x^2 + x^3)^n \) with respect to \( x \):
\[ \frac{d}{dx} \left( (1 + x + x^2 + x^3)^n \right) = n(1 + x + x^2 + x^3)^{n-1} \cdot (1 + 2x + 3x^2) \]
The right-hand side becomes:
\[ \sum_{r=0}^{3n} r b_r x^{r-1} \]

4. **Evaluate at \( x = 1 \)**:
Substitute \( x = 1 \) into the differentiated equation:
\[ n(1 + 1 + 1 + 1)^{n-1} \cdot (1 + 2 + 3) = n \cdot 4^{n-1} \cdot 6 \]
The left-hand side becomes:
\[ \sum_{r=0}^{3n} r b_r \]

5. **Combine the Results**:
Thus, we have:
\[ \sum_{r=0}^{3n} r b_r = n \cdot 4^{n-1} \cdot 6 \]

6. **Express in Terms of \( k \)**:
Since we know \( k = 4^n \), we can substitute:
\[ \sum_{r=0}^{3n} r b_r = n \cdot \frac{6k}{4} = \frac{3nk}{2} \]

### Final Result:
The final result for \( \sum_{r=0}^{3n} r b_r \) is:
\[ \sum_{r=0}^{3n} r b_r = \frac{3nk}{2} \]