Question

The first five terms of a quadratic sequence are 3, 12, 25, 42, 63.

Find the nth term of this sequence.

Answer 1

The nth term of the given quadratic sequence 3, 12, 25, 42, 63 is found by determining the pattern of the sequence and is calculated as 4n² + n + 3

Step-by-step explanation:

To find the nth term of a quadratic sequence, you first need to identify the pattern of the sequence given. The sequence provided is 3, 12, 25, 42, 63, which corresponds to the quadratic sequence rule, where the nth term is given by an equation of the form An² + Bn + C.

The differences between the terms are increasing by a constant amount, which is a common characteristic of a quadratic sequence. By examining the differences between the terms, we can begin to determine the coefficients A, B, and C that will give us our nth term formula.

Since we know that the differences are based on second differences being constant, and typically a quadratic sequence follows the pattern An² + Bn + C, we can establish that A is related to the second difference divided by 2. In this case, the second difference is consistent at 10, so A is 10/2 = 5. The sequence seems to follow the rule 5n², but we then adjust for the linear and constant terms by comparing and calculating the differences between the actual terms and the 5n² series.

After some calculations, which involve finding the linear and constant terms that align with the given sequence, one can deduce that the nth term of the given sequence is: 4n² + n + 3.

4 × 10² + 10 + 3 = 433

So, the 10th term of the sequence is 433.

Answer 2

Since we know the sequence is quadratic, we expect the
`n`-th term to have the general form

`x_n = an^2 + bn + c`

Plug in the first 3 known values of the sequence to form a system of equations.

`x_1 = 3 = a + b + c`

`x_2 = 12 = 4a + 2b + c`

`x_3 = 25 = 9a + 3b + c`

Eliminating
`c` gives

`(4a + 2b + c) - (a + b + c) = 12 - 3 \implies 3a + b = 9`

`(9a + 3b + c) - (a + b + c) = 25 - 3 \implies 8a + 2b = 22 \implies 4a + b = 11`

Eliminating
`b` gives

`(4a + b) - (3a + b) = 11 - 9 \implies a = 2`

Solving for
`b,c`, we get

`4a + b = 11 \implies b = 11-4\cdot2 = 3`

`a+b+c=3 \implies c=3-2-3 = -2`

So, the
`n`-th term of the sequence is

`x_n = \boxed{2n^2 + 3n - 2}`