If [x] denotes the greatest integer less then or equal to x,
then `[ (6sqrt(6) + 14)^(2n +1)]`

Options

A. is an even integer
B. is an odd integer
C. depands on n
D. none of these

Answer

To solve the problem of finding the greatest integer less than or equal to \( (6\sqrt{6} + 14)^{2n + 1} \), we can follow these steps:

### Step 1: Define the expression
Let:
\[ r = (6\sqrt{6} + 14)^{2n + 1} \]

### Step 2: Analyze the expression
We can express \( r \) as:
\[ r = (6\sqrt{6} + 14)^{2n + 1} = A + f \]
where \( A \) is the greatest integer less than or equal to \( r \) and \( f \) is the fractional part of \( r \) such that \( 0 \leq f < 1 \).

### Step 3: Consider the complementary expression
Now, consider the expression:
\[ s = (6\sqrt{6} - 14)^{2n + 1} \]
Since \( 6\sqrt{6} \) is approximately \( 14.7 \), \( s \) will be a small positive number when \( n \) is a non-negative integer.

### Step 4: Establish the relationship
From the definitions, we have:
\[ r - s = (6\sqrt{6} + 14)^{2n + 1} - (6\sqrt{6} - 14)^{2n + 1} \]
This difference will yield an even integer because the terms of odd powers will cancel out when expanded.

### Step 5: Simplify the expression
Thus, we can express:
\[ r = A + f \]
and since \( s \) is small, we can write:
\[ r - s = A + f - s \]
This means:
\[ r = A + f + s \]
where \( f + s \) is still less than 1.

### Step 6: Conclude about the greatest integer
Since \( r \) can be expressed as:
\[ r = A + (f + s) \]
and \( f + s < 1 \), we conclude that:
\[ \lfloor r \rfloor = A \]
which is an even integer since \( r - s \) is an even integer.

### Final Result
Thus, the greatest integer less than or equal to \( (6\sqrt{6} + 14)^{2n + 1} \) is an even integer.