In the expansion of `((1)/(x)+2x)^(n)` the sum of binomial coefficients is 6561, then the constant terin is

Options

A. `2^(4).""^(6)C_(4)`
B. `16.""^(8)C_(4)`
C. `""^(8)C_(4)`
D. none

Answer

To find the constant term in the expansion of \(\left(\frac{1}{x} + 2x\right)^n\) given that the sum of the binomial coefficients is 6561, we will follow these steps:

### Step 1: Find the value of \(n\)
The sum of the binomial coefficients in the expansion of \((a + b)^n\) is given by \(2^n\). In this case, we have:

\[ \text{Sum of coefficients} = \left(1 + 2\right)^n = 3^n \]

We are given that this sum is equal to 6561:

\[ 3^n = 6561 \]

Now, we need to express 6561 as a power of 3. We can calculate:

\[ 6561 = 3^8 \]

Thus, we can equate the exponents:

\[ n = 8 \]

### Step 2: Find the constant term in the expansion
The general term \(T_{r+1}\) in the expansion of \(\left(\frac{1}{x} + 2x\right)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} \left(\frac{1}{x}\right)^{n-r} (2x)^r \]

Substituting \(n = 8\):

\[ T_{r+1} = \binom{8}{r} \left(\frac{1}{x}\right)^{8-r} (2x)^r \]

This simplifies to:

\[ T_{r+1} = \binom{8}{r} \cdot 2^r \cdot \frac{1}{x^{8-r}} \cdot x^r = \binom{8}{r} \cdot 2^r \cdot \frac{1}{x^{8-2r}} \]

### Step 3: Set the exponent of \(x\) to zero for the constant term
For the term to be constant (independent of \(x\)), the exponent of \(x\) must be zero:

\[ 8 - 2r = 0 \]

Solving for \(r\):

\[ 2r = 8 \implies r = 4 \]

### Step 4: Substitute \(r\) back into the general term
Now we substitute \(r = 4\) back into the expression for the general term:

\[ T_{5} = \binom{8}{4} \cdot 2^4 \]

Calculating \(\binom{8}{4}\):

\[ \binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

Now calculate \(2^4\):

\[ 2^4 = 16 \]

Thus, the constant term is:

\[ T_{5} = 70 \cdot 16 = 1120 \]

### Final Answer
The constant term in the expansion is:

\[ \boxed{1120} \]