`(1+3/(1!) +9/(2!) +(27)/(3!) + .....oo) (1+9/(1!) +(81)/(2!)+(729)/(3!)+ .....oo)`=

Options

A. 1
B. `e^2`
C. `e^3`
D. `e^12`

Answer

To solve the given problem, we need to evaluate the product of two infinite series:

\[ (1 + \frac{3}{1!} + \frac{9}{2!} + \frac{27}{3!} + \ldots) \times (1 + \frac{9}{1!} + \frac{81}{2!} + \frac{729}{3!} + \ldots) \]

### Step 1: Identify the first series

The first series can be recognized as follows:

\[ 1 + \frac{3}{1!} + \frac{9}{2!} + \frac{27}{3!} + \ldots \]

This series resembles the Taylor series expansion for \( e^x \), which is:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

In our case, if we let \( x = 3 \), we can rewrite the first series as:

\[ e^3 \]

### Step 2: Identify the second series

Now, let's analyze the second series:

\[ 1 + \frac{9}{1!} + \frac{81}{2!} + \frac{729}{3!} + \ldots \]

This series also resembles the Taylor series expansion for \( e^x \). If we let \( x = 9 \), we can rewrite the second series as:

\[ e^9 \]

### Step 3: Multiply the two series

Now that we have identified both series, we can multiply them:

\[ e^3 \times e^9 \]

Using the property of exponents, we can combine the exponents:

\[ e^{3 + 9} = e^{12} \]

### Final Answer

Thus, the value of the given expression is:

\[ e^{12} \]