`e^(7x)-e^(-7x)=`

Options

A. `1 + (7x)/(2!) +((7x)^2)/(2!) +......`
B. `1 + ((7x^2))/(2!) +((7x)^4)/(4!) +......`
C. `2(1 + ((7x)^2)/(2!) +((7x)^4)/(4!) +......)`
D. `2(1 + (7x)/(1!) +((7x)^3)/(3!) +((7x)^5)/(5!) +...)`

Answer

To solve the expression \( e^{7x} - e^{-7x} \), we will use the Taylor series expansion for the exponential function. Here are the steps to find the solution:

### Step 1: Write the Taylor Series for \( e^{x} \)
The Taylor series expansion for \( e^{x} \) around \( x = 0 \) is given by:
\[ e^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

### Step 2: Substitute \( 7x \) into the Series
Now, substituting \( 7x \) into the series, we get:
\[ e^{7x} = 1 + \frac{7x}{1!} + \frac{(7x)^2}{2!} + \frac{(7x)^3}{3!} + \cdots \]

### Step 3: Write the Taylor Series for \( e^{-x} \)
Similarly, the Taylor series for \( e^{-x} \) is:
\[ e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \]

### Step 4: Substitute \( -7x \) into the Series
Substituting \( -7x \) into the series for \( e^{-x} \), we have:
\[ e^{-7x} = 1 - \frac{7x}{1!} + \frac{(7x)^2}{2!} - \frac{(7x)^3}{3!} + \cdots \]

### Step 5: Combine the Two Series
Now, we will subtract the series for \( e^{-7x} \) from the series for \( e^{7x} \):
\[ e^{7x} - e^{-7x} = \left(1 + \frac{7x}{1!} + \frac{(7x)^2}{2!} + \frac{(7x)^3}{3!} + \cdots\right) - \left(1 - \frac{7x}{1!} + \frac{(7x)^2}{2!} - \frac{(7x)^3}{3!} + \cdots\right) \]

### Step 6: Simplify the Expression
When we simplify this expression, we see that the constant terms (1) cancel out:
\[ e^{7x} - e^{-7x} = \left(\frac{7x}{1!} + \frac{(7x)^3}{3!} + \cdots\right) + \left(\frac{7x}{1!} + \frac{(7x)^3}{3!} + \cdots\right) \]
This results in:
\[ e^{7x} - e^{-7x} = 2\left(\frac{7x}{1!} + \frac{(7x)^3}{3!} + \cdots\right) \]

### Final Result
Thus, the final result is:
\[ e^{7x} - e^{-7x} = 2\left(\frac{7x}{1!} + \frac{(7x)^3}{3!} + \cdots\right) \]