The sum of the coefficients in the binomial expansion of `(1/x + 2x)^n` is equal to 6561. The constant term in the expansion is

Options

A. `""^8C_4`
B. `16(""^8C_4)`
C. `""^6C_4.2^4`
D. `8(""^8C_4)`

Answer

To solve the problem, we need to find the constant term in the binomial expansion of \((\frac{1}{x} + 2x)^n\) given that the sum of the coefficients is equal to 6561.

### Step-by-Step Solution:

1. **Find the value of \(n\)**:
- The sum of the coefficients in a binomial expansion can be found by substituting \(x = 1\).
- Therefore, we substitute \(x = 1\) into the expression:
\[ \left(\frac{1}{1} + 2 \cdot 1\right)^n = (1 + 2)^n = 3^n \]
- We know that this sum is equal to 6561:
\[ 3^n = 6561 \]
- Next, we express 6561 as a power of 3:
\[ 6561 = 3^8 \]
- Thus, we can equate the exponents:
\[ n = 8 \]

2. **Write the binomial expansion**:
- Now that we have \(n = 8\), we can write the expression as:
\[ \left(\frac{1}{x} + 2x\right)^8 \]

3. **Find the general term**:
- The general term \(T_{r+1}\) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} \left(\frac{1}{x}\right)^{n-r} (2x)^r \]
- Substituting \(n = 8\):
\[ T_{r+1} = \binom{8}{r} \left(\frac{1}{x}\right)^{8-r} (2x)^r \]
- Simplifying this, we get:
\[ T_{r+1} = \binom{8}{r} \cdot 2^r \cdot x^{r - (8 - r)} = \binom{8}{r} \cdot 2^r \cdot x^{2r - 8} \]

4. **Find the constant term**:
- The constant term occurs when the exponent of \(x\) is zero:
\[ 2r - 8 = 0 \]
- Solving for \(r\):
\[ 2r = 8 \implies r = 4 \]

5. **Substitute \(r\) back to find the constant term**:
- Now we substitute \(r = 4\) into the general term:
\[ T_{5} = \binom{8}{4} \cdot 2^4 \cdot x^{2 \cdot 4 - 8} \]
- This simplifies to:
\[ T_{5} = \binom{8}{4} \cdot 2^4 \cdot x^0 = \binom{8}{4} \cdot 16 \]
- Now, we calculate \(\binom{8}{4}\):
\[ \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]
- Thus, the constant term is:
\[ T_{5} = 70 \cdot 16 = 1120 \]

### Final Answer:
The constant term in the expansion is **1120**.