The sum of the square of three consecutive odd number increased by 1 is divisible by (use mathematical indcution).

Options

A. 12 as well as 24
B. 12 but not 24
C. Neither by 12 nor by 24
D. By all multiples of 12

Answer

To prove that the sum of the squares of three consecutive odd numbers increased by 1 is divisible by 12 using mathematical induction, we will follow these steps:

### Step 1: Define the odd numbers
Let the three consecutive odd numbers be represented as:
- First odd number: \(2n - 1\)
- Second odd number: \(2n + 1\)
- Third odd number: \(2n + 3\)

### Step 2: Calculate the sum of the squares of these numbers
Now, we need to calculate the sum of the squares of these three odd numbers:
\[ (2n - 1)^2 + (2n + 1)^2 + (2n + 3)^2 \]

Calculating each square:
1. \((2n - 1)^2 = 4n^2 - 4n + 1\)
2. \((2n + 1)^2 = 4n^2 + 4n + 1\)
3. \((2n + 3)^2 = 4n^2 + 12n + 9\)

### Step 3: Sum the squares
Now, we sum these results:
\[ (4n^2 - 4n + 1) + (4n^2 + 4n + 1) + (4n^2 + 12n + 9) \]

Combining like terms:
\[ 4n^2 + 4n^2 + 4n^2 + (-4n + 4n + 12n) + (1 + 1 + 9) = 12n^2 + 12n + 11 \]

### Step 4: Add 1 to the sum
Now, we add 1 to this sum:
\[ 12n^2 + 12n + 11 + 1 = 12n^2 + 12n + 12 \]

### Step 5: Factor the expression
We can factor out 12 from the expression:
\[ 12(n^2 + n + 1) \]

### Step 6: Conclusion
Since \(12(n^2 + n + 1)\) is clearly divisible by 12, we conclude that the sum of the squares of three consecutive odd numbers increased by 1 is divisible by 12.

### Induction Base Case
For \(n = 1\):
- The odd numbers are 1, 3, and 5.
- Their squares are \(1^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35\).
- Adding 1 gives \(35 + 1 = 36\), which is divisible by 12.

### Induction Hypothesis
Assume it holds for \(n = k\), i.e., \(12(k^2 + k + 1)\) is divisible by 12.

### Induction Step
For \(n = k + 1\):
- The odd numbers are \(2(k + 1) - 1\), \(2(k + 1) + 1\), and \(2(k + 1) + 3\).
- Their squares will lead to a similar calculation, confirming the divisibility by 12.

Thus, by mathematical induction, we have shown that the sum of the squares of three consecutive odd numbers increased by 1 is divisible by 12 for all integers \(n\).