Find the nth term of the quadratic sequences In the form n^2 + b 1) 8, 11, 16, 23, 32 2) 0, 3, 8, 15, 24               3) -8, -5, 0, 7, 16 In the form an^2 + b 1) 11, 17, 27, 41, 59...

Find the nth term of the quadratic sequences

In the form n^2 + b

1) 8, 11, 16, 23, 32

2) 0, 3, 8, 15, 24              

3) -8, -5, 0, 7, 16

In the form an^2 + b

1) 11, 17, 27, 41, 59

2) 7, 16, 31, 52, 79

3) 6, 18, 38, 66, 102

4) 0, 9, 24, 45, 72

5) 0, 12, 32, 60, 96

Asked on by eliza-15

2 Answers | Add Yours

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

Find the nth term of the quadratic sequences

The formulae for quadratic sequences are established by taking second differences between the list of terms. If the sequence is indeed quadratic (of order two) then the second differences will be constant and equal to `2a ` where `a ` is the coefficient of the quadratic term (term in `n^2 `).

We are told that the first set of sequences A1) - A3) are of the form `S_n = n^2 +b `  , where `S_n ` defines the `n `th term of the sequence, so that we should find that the second differences are constant and equal to twice the coefficient of the `n^2 ` term, namely 2.

So, for the first set of questions, take the (first) differences between the sequence terms and then the second differences (these are the differences between the sequence of first differences). Assume the first term of the sequence corresponds to  `n=1 `

A1) Sequence: 8, 11, 16, 23, 32

1st differences:  3, 5, 7, 9.  2nd differences: 2, 2, 2  (as expected)

Therefore `S_n = n^2 + 7 `

so that the value of the constant  `b `  is 7 (substitute values for `n `, ie 0, 1, 2, 3, 4 to see for yourself that this is true)

A2) Sequence: 0, 3, 8, 15, 24          

1st differences: 3, 5, 7, 9.  2nd differences: 2, 2, 2  

Therefore `S_n = n^2 -1 ` ` `

so that the value of the constant  `b `  is -1.

A3)  Sequence: -8, -5, 0, 7, 16

1st differences: 3, 5, 7, 9.  2nd differences: 2, 2, 2

Therefore `S_n = n^2 - 9 `

so that the value of the constant  `b `  is -9

-------------

For the second set of sequences B1) - B5) (the last two solutions follow directly from the first three so I will just give the solutions rather than working them through - after three examples I am confident you will get the idea) we are required to find the coefficient `a ` of the term in `n^2`   (the quadratic term) as well as the constant term  `b ` in the general sequence formula `S_n = an^2 + b`.  As before, take the first differences between the list of terms, and then the differences between those differences (ie the second differences). Also, again assume that the first term in the sequence corresponds to `n=1 `

B1)  Sequence: 11, 17, 27, 41, 59

1st differences: 6, 10, 14, 18.  2nd differences:  4, 4, 4

Firstly we notice (from the second differences) that `2a = 4 `  so that we have

`S_n = 2n^2 + b ` ` `

Substituting `n=1 ` into this formula and matching to the first term in the original sequence we get `b = 11 - 2 = 9 ` so that

`S_n = 2n^2 + 9 `

B2) Sequence: 7, 16, 31, 52, 79

1st differences: 9, 15, 21, 27.  2nd differences: 6, 6, 6

Firstly we notice (from the second differences) that  `2a = 6 `  so that we have

`S_n = 3n^2 + b `

` `

Substituting ` ` `n=1 ` into this formula and matching to the first term in the original sequence we get `b = 7 - 3 = 4 ` so that

`S_n = 3n^2 + 4 `

` `

B3) Sequence: 6, 18, 38, 66, 102

1st differences: 12, 20, 28, 36.  2nd differences: 8, 8, 8

Firstly we notice (from the second differences) that  `2a = 8 ` ` ` so that we have

`S_n = 4n^2 + b `

Substituting `n=1 ` into this formula and matching to the first term in the sequence we get `b = 6 - 4 = 2 `  so that

`S_n = 4n^2 + 2 `

` `

B4) Sequence: 0, 9, 24, 45, 72.  Formula for nth term:

`S_n = 3n^2 - 3 `

B5) Sequence: 0, 12, 32, 60, 96. Formula for nth term:

`````S_n = 4n^2 - 4 `

` `` `

Sources:
steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

To find the first group, you need to find b.  You start with plugging in n = 1.  So, for the first sequence 8, 11, 16, 23, 32, we would have:

1^2 + b = 8

1 + b = 8

b = 7

This works for n = 2, also:

2^2 + b = 11

4 + b = 11

b = 7

So, for the first sequence, we have:

n^2 + 7

Following the same procedure for the second sequence, we have:

n^2 + (-1)

For the third sequence:

n^2 + (-9)

For the second set of sequences, we have to solve a system of equations.  Each equation comes from plugging in the sequential number for n.  For example, for the first sequence 11, 17, 27, 41, 59, we have:

n = 1; a*1^2 + b = 11

a + b = 11

n = 2; a*2^2 + b = 17

4a + b = 17

So we have the system of equations:

4a+b = 17

a+b = 11

Solving this system of equations, we get a = 2 and b = 9.  So, the sequence for the first one in the second set is 2n^2 + 9.

Using the same procedure for the other sequences, we get:

2) 3n^2 + 4

3) 4n^2 + 2

4) 3n^2 + (-3)

5) 4n^2 + (-4)

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question