The value of
`(1 + 3) log 3 + ((1+3^(2)))/(2!) (log 3)^(2) + ((1+3^(3)))/(3!) (log 3)^(3) + ...oo` is

Options

A. 18
B. 28
C. 36
D. none

Answer

To solve the given expression

\[ (1 + 3) \log 3 + \frac{(1 + 3^2)}{2!} (\log 3)^2 + \frac{(1 + 3^3)}{3!} (\log 3)^3 + \ldots \]

let's denote the entire expression as \( I \).

### Step 1: Rewrite the expression

We can rewrite \( I \) as:

\[ I = 4 \log 3 + \frac{(1 + 9)}{2!} (\log 3)^2 + \frac{(1 + 27)}{3!} (\log 3)^3 + \ldots \]

This simplifies to:

\[ I = 4 \log 3 + \frac{10}{2!} (\log 3)^2 + \frac{28}{3!} (\log 3)^3 + \ldots \]

### Step 2: Separate the terms

Now, we can separate the terms into two series:

1. The series involving \( \log 3 \):
\[ S_1 = \log 3 + \frac{(\log 3)^2}{2!} + \frac{(\log 3)^3}{3!} + \ldots \]

2. The series involving \( 3 \log 3 \):
\[ S_2 = 3 \log 3 + \frac{9 (\log 3)^2}{2!} + \frac{27 (\log 3)^3}{3!} + \ldots \]

### Step 3: Recognize the series

The first series \( S_1 \) can be recognized as the Taylor series expansion for \( e^x - 1 \) evaluated at \( x = \log 3 \):

\[ S_1 = e^{\log 3} - 1 = 3 - 1 = 2 \]

The second series \( S_2 \) can be recognized similarly, but with \( x = 3 \log 3 \):

\[ S_2 = e^{3 \log 3} - 1 = e^{\log 27} - 1 = 27 - 1 = 26 \]

### Step 4: Combine the results

Now, we can combine the results from \( S_1 \) and \( S_2 \):

\[ I = 4 \cdot 2 + 26 = 8 + 26 = 34 \]

### Step 5: Final result

Thus, the value of the given expression is

\[ \boxed{34} \]

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