The expansion of `(x^(alpha)+1/x^(beta))^(n)` has constant term, if

Options

A. `n alpha` is divisible by `n+beta`
B. `nbeta` is divisible by `n+alpha`
C. `nalpha` is divisible by `alpha+beta`
D. n is divisible by `alpha+beta`

Answer

To find the conditions under which the expansion of \((x^{\alpha} + \frac{1}{x^{\beta}})^{n}\) has a constant term, we can follow these steps:

### Step 1: Write the General Term of the Expansion
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression, let \(a = x^{\alpha}\) and \(b = \frac{1}{x^{\beta}}\). Thus, the general term becomes:
\[ T_r = \binom{n}{r} (x^{\alpha})^{n-r} \left(\frac{1}{x^{\beta}}\right)^r \]

### Step 2: Simplify the General Term
Now, simplifying \(T_r\):
\[ T_r = \binom{n}{r} x^{\alpha(n-r)} \cdot x^{-\beta r} = \binom{n}{r} x^{\alpha(n-r) - \beta r} \]
This can be rewritten as:
\[ T_r = \binom{n}{r} x^{n\alpha - r(\alpha + \beta)} \]

### Step 3: Set the Exponent of \(x\) to Zero for the Constant Term
For \(T_r\) to be a constant term, the exponent of \(x\) must be zero:
\[ n\alpha - r(\alpha + \beta) = 0 \]

### Step 4: Solve for \(r\)
Rearranging the equation gives:
\[ n\alpha = r(\alpha + \beta) \]
Thus, we can solve for \(r\):
\[ r = \frac{n\alpha}{\alpha + \beta} \]

### Step 5: Ensure \(r\) is an Integer
For \(r\) to be a valid term in the binomial expansion, it must be an integer. Therefore, \(n\alpha\) must be divisible by \(\alpha + \beta\):
\[ n\alpha \text{ is divisible by } (\alpha + \beta) \]

### Conclusion
The expansion of \((x^{\alpha} + \frac{1}{x^{\beta}})^{n}\) has a constant term if and only if \(n\alpha\) is divisible by \(\alpha + \beta\).

---