If `(e^(x) +e^(5x))/( e^(3x)) = a_(0) +a_(1)x +a_(2)x^(2) +a_(3)x^(3) +….` then the value of `2a_(1) +2^(3) a_(3) +2^(5) a_(5)+ …… ` is

Options

A. `e^(2) +e^(-2)`
B. `e^(4)-e^(-4)`
C. `e^(4)+e^(-4)`
D. 0

Answer

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

\[ \frac{e^x + e^{5x}}{e^{3x}} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \]

### Step 1: Simplify the left-hand side

We can rewrite the left-hand side as:

\[ \frac{e^x}{e^{3x}} + \frac{e^{5x}}{e^{3x}} = e^{-2x} + e^{2x} \]

### Step 2: Expand both terms using Taylor series

The Taylor series expansion for \(e^{kx}\) is given by:

\[ e^{kx} = \sum_{n=0}^{\infty} \frac{(kx)^n}{n!} \]

Applying this to both terms:

1. For \(e^{-2x}\):

\[ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = 1 - 2x + \frac{(-2x)^2}{2!} - \frac{(-2x)^3}{3!} + \ldots \]

2. For \(e^{2x}\):

\[ e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \ldots \]

### Step 3: Combine the series

Now we combine the two series:

\[ e^{-2x} + e^{2x} = \left(1 - 2x + \frac{4x^2}{2} - \frac{(-8x^3)}{6} + \ldots\right) + \left(1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \ldots\right) \]

### Step 4: Identify the coefficients

When we combine these series, we notice:

- The constant terms \(1 + 1 = 2\) gives \(a_0 = 2\).
- The coefficients of \(x\) cancel out: \(-2x + 2x = 0\) gives \(a_1 = 0\).
- The coefficients of \(x^2\) add up: \(\frac{4x^2}{2} + \frac{4x^2}{2} = 4\) gives \(a_2 = 4\).
- The coefficients of \(x^3\) cancel out: \(-\frac{8x^3}{6} + \frac{8x^3}{6} = 0\) gives \(a_3 = 0\).
- Continuing this pattern, we find that all odd-indexed coefficients \(a_1, a_3, a_5, \ldots\) are zero.

### Step 5: Calculate the required sum

We need to calculate:

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

Since \(a_1 = 0\), \(a_3 = 0\), \(a_5 = 0\), and so on, we find:

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

### Final Answer

Thus, the value of \(2a_1 + 2^3 a_3 + 2^5 a_5 + \ldots\) is:

\[ \boxed{0} \]