If the sum of the coefficients in the expansion of `(1+3x)^(n)` lies between 4000 and 10000, then the value of the greatest coefficient must be

Options

A. 3954
B. 6342
C. 4806
D. 1458

Answer

To solve the problem step by step, we need to find the greatest coefficient in the expansion of \((1 + 3x)^n\) given that the sum of the coefficients lies between 4000 and 10000.

### Step 1: Find the sum of the coefficients
The sum of the coefficients in the expansion of \((1 + 3x)^n\) can be found by substituting \(x = 1\):
\[ (1 + 3 \cdot 1)^n = (1 + 3)^n = 4^n \]
So, the sum of the coefficients is \(4^n\).

### Step 2: Set up the inequality
We know that the sum of the coefficients lies between 4000 and 10000:
\[ 4000 < 4^n < 10000 \]

### Step 3: Solve the inequalities
To find \(n\), we can take logarithms or evaluate powers of 4.

- First, calculate \(4^n > 4000\):
- \(4^n = 4^6 = 4096\) (this is greater than 4000)
- \(4^5 = 1024\) (this is less than 4000)

Thus, \(n\) must be at least 6.

- Now, calculate \(4^n < 10000\):
- \(4^7 = 16384\) (this is greater than 10000)
- \(4^6 = 4096\) (this is less than 10000)

Thus, \(n\) can be at most 6.

From the above, we conclude that:
\[ n = 6 \]

### Step 4: Find the greatest coefficient
The greatest coefficient in the expansion of \((1 + 3x)^n\) occurs at the term where \(k\) is approximately \(\frac{n}{2}\). For \(n = 6\), we calculate:
- The coefficients are given by \(\binom{n}{k} (3x)^k\).

The greatest coefficient occurs at:
\[ k = \left\lfloor \frac{n}{2} \right\rfloor = 3 \quad \text{or} \quad k = \left\lceil \frac{n}{2} \right\rceil = 3 \]

The coefficient for \(k = 3\) is:
\[ \binom{6}{3} (3)^3 \]

Calculating this:
\[ \binom{6}{3} = 20 \quad \text{and} \quad (3)^3 = 27 \]
Thus, the coefficient is:
\[ 20 \cdot 27 = 540 \]

### Final Answer
The value of the greatest coefficient must be:
\[ \boxed{540} \]