If `x =3 + (4)/(2!) + (8)/(3!) + (16)/(4!) + .....oo ` then `1/x =`

Options

A. `1/e`
B. e
C. `e^2`
D. `1/e^2`

Answer

To solve the problem step by step, we start with the given series for \( x \):

### Step 1: Write the series
The series is given as:
\[ x = 3 + \frac{4}{2!} + \frac{8}{3!} + \frac{16}{4!} + \ldots \]

### Step 2: Rewrite the terms
Notice that the terms can be rewritten as follows:
- The first term \( 3 \) can be expressed as \( 1 + 2 \).
- The second term \( \frac{4}{2!} \) can be rewritten as \( \frac{2^2}{2!} \).
- The third term \( \frac{8}{3!} \) can be rewritten as \( \frac{2^3}{3!} \).
- The fourth term \( \frac{16}{4!} \) can be rewritten as \( \frac{2^4}{4!} \).

Thus, we can express \( x \) as:
\[ x = 1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots \]

### Step 3: Identify the series
The series \( 1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots \) resembles the Taylor series expansion for \( e^x \), specifically:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \]
If we let \( x = 2 \), we get:
\[ e^2 = 1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \ldots \]

### Step 4: Conclude the value of \( x \)
From the above, we can conclude that:
\[ x = e^2 \]

### Step 5: Find \( \frac{1}{x} \)
Now, we need to find \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{1}{e^2} \]

### Step 6: Simplify the expression
Using the property of exponents, we can express this as:
\[ \frac{1}{x} = e^{-2} \]

### Final Answer
Thus, the value of \( \frac{1}{x} \) is:
\[ \frac{1}{x} = e^{-2} \]