QUESTION 2 Investigate the difference (D) between (the sum of the first and third terms) and (two times the second/middle term) of any consecutive numbers of a quadratic sequence. Glven: $$ (1,3 ; 6 ; \ldots 12 ; 4 ; 8 ; \ldots 2 ; 6 ; 12 ; 20 ; \ldots 13 ; 4 ; 8 ; 15 ; \ldots) $$ NOTE: If the sequence is given by $$ a ; b ; c ; \ldots $$ the difference $$ (D)=(a+c)-2 imes b $$. Step 1 - Use the given sequences and determine the Difference, D, in each case. Step 2 - Compare your answers in Step 1 with the second difference for each sequence. Step 3 -What is your conjecture? Step 4 - Prove that this is TRUE in general.

Question

QUESTION 2 Investigate the difference (D) between (the sum of the first and third terms) and (two times the second/middle term) of any consecutive numbers of a quadratic sequence. Glven: $$ (1,3 ; 6 ; \ldots 12 ; 4 ; 8 ; \ldots 2 ; 6 ; 12 ; 20 ; \ldots 13 ; 4 ; 8 ; 15 ; \ldots) $$ NOTE: If the sequence is given by $$ a ; b ; c ; \ldots $$ the difference $$ (D)=(a+c)-2 	imes b $$. Step 1 - Use the given sequences and determine the Difference, D, in each case. Step 2 - Compare your answers in Step 1 with the second difference for each sequence. Step 3 -What is your conjecture? Step 4 - Prove that this is TRUE in general.

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the question.

### Step 1: Calculate the Difference $$ D $$

We will analyze the given sequences and calculate the difference $$ D = (a + c) - 2b $$ for each sequence.

1. **Sequence 1: $$ 1, 3, 6 $$**
   - $$ a = 1 $$
   - $$ b = 3 $$
   - $$ c = 6 $$
   - Calculate $$ D $$:
     $$
     D = (1 + 6) - 2 \times 3 = 7 - 6 = 1
     $$

2. **Sequence 2: $$ 4, 8, 12 $$**
   - $$ a = 4 $$
   - $$ b = 8 $$
   - $$ c = 12 $$
   - Calculate $$ D $$:
     $$
     D = (4 + 12) - 2 \times 8 = 16 - 16 = 0
     $$

3. **Sequence 3: $$ 2, 6, 12 $$**
   - $$ a = 2 $$
   - $$ b = 6 $$
   - $$ c = 12 $$
   - Calculate $$ D $$:
     $$
     D = (2 + 12) - 2 \times 6 = 14 - 12 = 2
     $$

4. **Sequence 4: $$ 20, 13, 4 $$**
   - $$ a = 20 $$
   - $$ b = 13 $$
   - $$ c = 4 $$
   - Calculate $$ D $$:
     $$
     D = (20 + 4) - 2 \times 13 = 24 - 26 = -2
     $$

### Summary of Differences $$ D $$:
- Sequence 1: $$ D = 1 $$
- Sequence 2: $$ D = 0 $$
- Sequence 3: $$ D = 2 $$
- Sequence 4: $$ D = -2 $$

### Step 2: Compare with the Second Difference

To find the second difference, we first need to calculate the first differences of each sequence.

1. **Sequence 1: $$ 1, 3, 6 $$**
   - First differences: $$ 3 - 1 = 2 $$, $$ 6 - 3 = 3 $$
   - Second difference: $$ 3 - 2 = 1 $$

2. **Sequence 2: $$ 4, 8, 12 $$**
   - First differences: $$ 8 - 4 = 4 $$, $$ 12 - 8 = 4 $$
   - Second difference: $$ 4 - 4 = 0 $$

3. **Sequence 3: $$ 2, 6, 12 $$**
   - First differences: $$ 6 - 2 = 4 $$, $$ 12 - 6 = 6 $$
   - Second difference: $$ 6 - 4 = 2 $$

4. **Sequence 4: $$ 20, 13, 4 $$**
   - First differences: $$ 13 - 20 = -7 $$, $$ 4 - 13 = -9 $$
   - Second difference: $$ -9 - (-7) = -2 $$

### Summary of Second Differences:
- Sequence 1: Second difference = $$ 1 $$
- Sequence 2: Second difference = $$ 0 $$
- Sequence 3: Second difference = $$ 2 $$
- Sequence 4: Second difference = $$ -2 $$

### Step 3: Conjecture

From the calculations, we can observe that the difference $$ D $$ calculated in Step 1 matches the second difference calculated in Step 2 for each sequence. Thus, we can conjecture that:

**Conjecture:** The difference $$ D $$ between the sum of the first and third terms and two times the second term of any quadratic sequence is equal to the second difference of that sequence.

### Step 4: Prove that this is TRUE in general

To prove this conjecture, we can consider a general quadratic sequence defined by the $$ n $$-th term:

$$
T_n = An^2 + Bn + C
$$

Where $$ A, B, C $$ are constants.

1. **First term:** $$ T_1 = A(1)^2 + B(1) + C = A + B + C $$
2. **Second term:** $$ T_2 = A(2)^2 + B(2) + C = 4A + 2B + C $$
3. **Third term:** $$ T_3 = A(3)^2 + B(3) + C = 9A + 3B + C $$

Now, calculate $$ D $$:

$$
D = (T_1 + T_3) - 2T_2
$$
$$
D = (A + B + C + 9A + 3B + C) - 2(4A + 2B + C)
$$
$$
D = (10A + 4B + 2C) - (8A + 4B + 2C)
$$
$$
D = 2A
$$

Next, we calculate the first differences:

- First difference $$ T_2 - T_1 = (4A + 2B + C) - (A + B + C) = 3A + B $$
- First difference $$ T_3 - T_2 = (9A + 3B + C) - (4A + 2B + C) = 5A + B $$

Now, the second difference:

$$
\text{Second difference} = (5A + B) - (3A + B) = 2A
$$

Thus, we have shown that:

$$
D = 2A = \text{Second difference}
$$

This proves that the conjecture is true in general.
The difference $$ D $$ between the sum of the first and third terms and two times the second term of any quadratic sequence is equal to the second difference of that sequence.

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