If sum of the coefficients in the expansion of `(x+1/x)^(n)` is 128, then coefficient of x in the expansion of `(x+1/x)^(n)` is

Options

A. 51
B. 48
C. 35
D. 28

Answer

To solve the problem, we need to find the coefficient of \( x \) in the expansion of \( (x + \frac{1}{x})^n \) given that the sum of the coefficients is 128. Here’s a step-by-step solution:

### Step 1: Understanding the sum of coefficients
The sum of the coefficients in the expansion of \( (x + \frac{1}{x})^n \) can be found by substituting \( x = 1 \) and \( \frac{1}{x} = 1 \). Thus, we have:
\[ (1 + 1)^n = 2^n \]
Given that this sum is equal to 128, we can set up the equation:
\[ 2^n = 128 \]

### Step 2: Solving for \( n \)
Next, we need to solve for \( n \). We know that:
\[ 128 = 2^7 \]
Thus, we can equate the exponents:
\[ n = 7 \]

### Step 3: Finding the coefficient of \( x \)
Now we need to find the coefficient of \( x \) in the expansion of \( (x + \frac{1}{x})^7 \). The general term in the expansion is given by:
\[ \binom{n}{r} x^r \left(\frac{1}{x}\right)^{n-r} = \binom{n}{r} x^{r - (n - r)} = \binom{n}{r} x^{2r - n} \]
For \( n = 7 \), this becomes:
\[ \binom{7}{r} x^{2r - 7} \]
We need the power of \( x \) to be 1, so we set:
\[ 2r - 7 = 1 \]

### Step 4: Solving for \( r \)
Now, we solve for \( r \):
\[ 2r = 8 \implies r = 4 \]

### Step 5: Finding the coefficient for \( r = 4 \)
Now we substitute \( r = 4 \) into the binomial coefficient:
\[ \text{Coefficient of } x = \binom{7}{4} \]
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
we calculate:
\[ \binom{7}{4} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \]

### Final Answer
Thus, the coefficient of \( x \) in the expansion of \( (x + \frac{1}{x})^7 \) is:
\[ \boxed{35} \]