If `(1)/(e^(3x))(e^(x)+e^(5x))=a^(0)+a_(1)x+a_(2)x^(2)+ . . .,` then `2a_(1)+2^(3)a_(3)+2^(5)a_5+ . . . ` is equal to

Options

A. e
B. `e^(-1)`
C. 1
D. 0

Answer

To solve the given problem, we start with the equation:

\[ \frac{1}{e^{3x}(e^{x} + e^{5x})} \]

We can simplify this expression. First, we rewrite it:

\[ \frac{1}{e^{3x}(e^{x} + e^{5x})} = \frac{1}{e^{3x}} \cdot \frac{1}{e^{x} + e^{5x}} \]

Next, we factor out \( e^{x} \) from the denominator:

\[ e^{x} + e^{5x} = e^{x}(1 + e^{4x}) \]

Thus, we can rewrite the expression as:

\[ \frac{1}{e^{3x}} \cdot \frac{1}{e^{x}(1 + e^{4x})} = \frac{1}{e^{4x}(1 + e^{4x})} \]

Now, we can further simplify this:

\[ \frac{1}{e^{4x}(1 + e^{4x})} = \frac{1}{e^{4x}} \cdot \frac{1}{1 + e^{4x}} \]

Now, we can express this in terms of a power series. The term \( \frac{1}{1 + e^{4x}} \) can be expanded using the geometric series formula:

\[ \frac{1}{1 + e^{4x}} = \sum_{n=0}^{\infty} (-1)^n (e^{4x})^n = \sum_{n=0}^{\infty} (-1)^n e^{4nx} \]

Therefore, we have:

\[ \frac{1}{e^{4x}(1 + e^{4x})} = e^{-4x} \sum_{n=0}^{\infty} (-1)^n e^{4nx} = \sum_{n=0}^{\infty} (-1)^n e^{(4n - 4)x} \]

Now, we want to find the coefficients \( a_n \) in the series expansion:

\[ \sum_{n=0}^{\infty} a_n x^n \]

From the above expression, we can see that \( a_n \) corresponds to the coefficients of \( x^n \) in the expansion. Specifically, we need to find \( 2a_1 + 2^3 a_3 + 2^5 a_5 + \ldots \).

To find \( a_n \), we note that \( a_n \) is non-zero only for \( n = 4k \) (where \( k \) is a non-negative integer), and the coefficients alternate in sign. Thus we have:

\[ a_{4k} = (-1)^k \]

Now we compute \( 2a_1 + 2^3 a_3 + 2^5 a_5 + \ldots \):

1. For \( n = 1 \): \( a_1 = 0 \)
2. For \( n = 3 \): \( a_3 = 0 \)
3. For \( n = 5 \): \( a_5 = 0 \)
4. Continuing this, we find that \( a_n = 0 \) for all odd \( n \).

Thus:

\[ 2a_1 + 2^3 a_3 + 2^5 a_5 + \ldots = 0 \]

Therefore, the final answer is:

\[ \boxed{0} \]