Suppose `a in R`. If the coefficient of `x^(5)` in the expansion of `(ax+1/(x^(3)))^(17)` is 680, then a is equal to

Options

A. `pm2`
B. `pm1`
C. `pm1//2`
D. `pm1//3`

Answer

To solve the problem, we need to find the value of \( a \) such that the coefficient of \( x^5 \) in the expansion of \( (ax + \frac{1}{x^3})^{17} \) is equal to 680.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (A + B)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} A^{n-r} B^r \]
Here, \( A = ax \) and \( B = \frac{1}{x^3} \), and \( n = 17 \).

2. **Write the General Term for Our Expression**:
Thus, the general term for our expression becomes:
\[ T_{r+1} = \binom{17}{r} (ax)^{17-r} \left(\frac{1}{x^3}\right)^r \]
Simplifying this gives:
\[ T_{r+1} = \binom{17}{r} a^{17-r} x^{17-r} \cdot x^{-3r} = \binom{17}{r} a^{17-r} x^{17 - 4r} \]

3. **Set the Power of \( x \) to 5**:
We need the coefficient of \( x^5 \). Therefore, we set:
\[ 17 - 4r = 5 \]

4. **Solve for \( r \)**:
Rearranging the equation gives:
\[ 17 - 5 = 4r \implies 12 = 4r \implies r = 3 \]

5. **Substitute \( r \) Back into the General Term**:
Now substitute \( r = 3 \) into the general term to find the coefficient:
\[ T_{4} = \binom{17}{3} a^{14} \]

6. **Calculate \( \binom{17}{3} \)**:
The binomial coefficient \( \binom{17}{3} \) can be calculated as:
\[ \binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = \frac{4080}{6} = 680 \]

7. **Set Up the Equation**:
Therefore, the coefficient of \( x^5 \) is:
\[ 680 \cdot a^{14} = 680 \]

8. **Solve for \( a^{14} \)**:
Dividing both sides by 680 gives:
\[ a^{14} = 1 \]

9. **Find the Values of \( a \)**:
The solutions to \( a^{14} = 1 \) are:
\[ a = 1 \quad \text{or} \quad a = -1 \]

### Final Answer:
Thus, the possible values of \( a \) are:
\[ \boxed{1 \text{ or } -1} \]