`3^(11)+3^(12)+3^(13)+3^(14)` is divisible by

Options

A. 7
B. 8
C. 11
D. 14

Answer

To solve the expression \(3^{11} + 3^{12} + 3^{13} + 3^{14}\) and determine its divisibility, we can follow these steps:

### Step 1: Factor out the common term
We notice that all terms have a common factor of \(3^{11}\). Therefore, we can factor it out:

\[ 3^{11} + 3^{12} + 3^{13} + 3^{14} = 3^{11}(1 + 3 + 3^2 + 3^3) \]

### Step 2: Simplify the expression inside the parentheses
Now, we simplify the expression inside the parentheses:

\[ 1 + 3 + 3^2 + 3^3 = 1 + 3 + 9 + 27 \]

Calculating this step-by-step:

- \(1 + 3 = 4\)
- \(4 + 9 = 13\)
- \(13 + 27 = 40\)

So, we have:

\[ 1 + 3 + 3^2 + 3^3 = 40 \]

### Step 3: Substitute back into the factored expression
Now, substituting back into our factored expression gives us:

\[ 3^{11} \times 40 \]

### Step 4: Analyze the divisibility
Now we need to determine what \(3^{11} \times 40\) is divisible by.

- \(3^{11}\) is a power of 3, which is not divisible by 2.
- \(40\) can be factored as \(2^3 \times 5\).

Thus, \(3^{11} \times 40\) is divisible by \(40\), which is \(8\) (since \(2^3 = 8\)) and \(5\).

### Conclusion
The expression \(3^{11} + 3^{12} + 3^{13} + 3^{14}\) is divisible by \(40\).