Question

Find the first five terms of the quadratic polynomial sequence with the formula `a_{n}=3n^{2}-2`. Calculate the 1st and 2nd differences.

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Answer 1

To find the first five terms of the quadratic polynomial sequence given by the formula \( a_n = 3n^2 - 2 \), we will substitute the values of \( n \) from 1 to 5 into the formula.

### Step 1: Calculate the First Five Terms

1. For \( n = 1 \):
\[
a_1 = 3(1^2) - 2 = 3(1) - 2 = 3 - 2 = 1
\]

2. For \( n = 2 \):
\[
a_2 = 3(2^2) - 2 = 3(4) - 2 = 12 - 2 = 10
\]

3. For \( n = 3 \):
\[
a_3 = 3(3^2) - 2 = 3(9) - 2 = 27 - 2 = 25
\]

4. For \( n = 4 \):
\[
a_4 = 3(4^2) - 2 = 3(16) - 2 = 48 - 2 = 46
\]

5. For \( n = 5 \):
\[
a_5 = 3(5^2) - 2 = 3(25) - 2 = 75 - 2 = 73
\]

### The first five terms of the sequence are:
\[
1, 10, 25, 46, 73
\]

### Step 2: Calculate the First Differences

The first differences are calculated by subtracting each term from the subsequent term:

- \( a_2 - a_1 = 10 - 1 = 9 \)
- \( a_3 - a_2 = 25 - 10 = 15 \)
- \( a_4 - a_3 = 46 - 25 = 21 \)
- \( a_5 - a_4 = 73 - 46 = 27 \)

### The first differences are:
\[
9, 15, 21, 27
\]

### Step 3: Calculate the Second Differences

The second differences are calculated by subtracting each first difference from the subsequent first difference:

- \( 15 - 9 = 6 \)
- \( 21 - 15 = 6 \)
- \( 27 - 21 = 6 \)

### The second differences are:
\[
6, 6, 6
\]

### Summary
- The first five terms of the quadratic polynomial sequence are: \( 1, 10, 25, 46, 73 \)
- The first differences are: \( 9, 15, 21, 27 \)
- The second differences are: \( 6, 6, 6 \)

Thus, we can conclude that in a quadratic sequence, the second differences are constant, confirming that our sequence is indeed quadratic.