If `(e^(5x)+e^(x))/(e^(3x))` is expand in a series of ascending powers of x and n is and odd natural number then the coefficent of `x^(n)` is

Options

A. `(2^(n))/(n!)`
B. `(2^(n+1))/(2n)!`
C. `(2^(2n))/(2n)!`
D. none of these

Answer

To solve the problem, we need to expand the expression \((e^{5x} + e^{x}) / e^{3x}\) in a series of ascending powers of \(x\) and find the coefficient of \(x^n\) where \(n\) is an odd natural number.

### Step 1: Expand \(e^{5x}\), \(e^{x}\), and \(e^{3x}\)
Using the Taylor series expansion for \(e^x\), we have:
\[ e^{kx} = 1 + kx + \frac{(kx)^2}{2!} + \frac{(kx)^3}{3!} + \cdots \]
Thus, we can write:
\[ e^{5x} = 1 + 5x + \frac{(5x)^2}{2!} + \frac{(5x)^3}{3!} + \cdots \]
\[ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
\[ e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + \cdots \]

### Step 2: Combine the Numerator
Now, we combine \(e^{5x}\) and \(e^{x}\):
\[ e^{5x} + e^{x} = \left(1 + 5x + \frac{(5x)^2}{2!} + \frac{(5x)^3}{3!} + \cdots\right) + \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) \]
Combining like terms:
\[ = 2 + (5x + x) + \left(\frac{(5^2)x^2}{2!} + \frac{x^2}{2!}\right) + \left(\frac{(5^3)x^3}{3!} + \frac{x^3}{3!}\right) + \cdots \]
\[ = 2 + 6x + \left(\frac{25 + 1}{2}x^2\right) + \left(\frac{125 + 1}{6}x^3\right) + \cdots \]

### Step 3: Expand the Denominator
Next, we expand the denominator \(e^{3x}\):
\[ e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + \cdots \]

### Step 4: Form the Complete Expression
Now we have:
\[ \frac{e^{5x} + e^{x}}{e^{3x}} = \frac{2 + 6x + \frac{26}{2}x^2 + \frac{126}{6}x^3 + \cdots}{1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 + \cdots} \]

### Step 5: Coefficient of \(x^n\)
To find the coefficient of \(x^n\) where \(n\) is an odd natural number, we note that the numerator contributes terms of the form \(a_n x^n\) and the denominator contributes terms of the form \(b_n x^n\). The coefficient of \(x^n\) in the quotient can be found using the formula for coefficients in series division.

However, since we are looking for odd \(n\), we can observe that the numerator will have contributions from odd powers of \(x\) and the denominator will have contributions from even powers of \(x\). Thus, when we divide, the odd powers will not yield a non-zero coefficient for odd \(n\).

### Conclusion
Therefore, the coefficient of \(x^n\) for odd \(n\) in the series expansion of \((e^{5x} + e^{x}) / e^{3x}\) is zero.

### Final Answer:
The coefficient of \(x^n\) where \(n\) is an odd natural number is **0**.