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.

Answer 1

To find the first five terms of the quadratic polynomial sequence given by the formula \( a_n = 3n^2 - 2 \), we will evaluate it for \( n = 1, 2, 3, 4, \) and \( 5 \).

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

Now we have the first five terms of the sequence:
\[
1, 10, 25, 46, 73
\]

Next, we calculate the first differences. The first difference \( \Delta_1 \) is calculated as \( a_{n+1} - a_n \):

- \( \Delta_1(1) = a_2 - a_1 = 10 - 1 = 9 \)
- \( \Delta_1(2) = a_3 - a_2 = 25 - 10 = 15 \)
- \( \Delta_1(3) = a_4 - a_3 = 46 - 25 = 21 \)
- \( \Delta_1(4) = a_5 - a_4 = 73 - 46 = 27 \)

So, the first differences are:
\[
9, 15, 21, 27
\]

Next, we will calculate the second differences \( \Delta_2 \) which are the differences of the first differences:

- \( \Delta_2(1) = \Delta_1(2) - \Delta_1(1) = 15 - 9 = 6 \)
- \( \Delta_2(2) = \Delta_1(3) - \Delta_1(2) = 21 - 15 = 6 \)
- \( \Delta_2(3) = \Delta_1(4) - \Delta_1(3) = 27 - 21 = 6 \)

Thus, the second differences are:
\[
6, 6, 6
\]

To summarize:
- The first five terms of the sequence are \( 1, 10, 25, 46, 73 \).
- The first differences are \( 9, 15, 21, 27 \).
- The second differences are \( 6, 6, 6 \).