Quadratic Sequences

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Warm-Up:

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Decribing a Quadratic Sequence:

A quadratic sequence has $n^2$ and no higher powers of $n$ in its $n^{\mathrm{th}}$ term expression ($U_n$). The general form of a quadratic sequence is given by

$$U_n=a{n}^2+bn+c$$

where $a,b$ and $c$ are numbers that can be thought of as characterising the quadratic sequence and often are to be found.

An example of a quadratic sequence is $1,4,9,16,25,...$ which happens to be the square numbers. Another is $3,6,13,24,39,...$ which is less easily identified as a particular sequence. To determine if a given sequence, is indeed, a quadratic sequence the difference between terms needs to be shown to be an arithmetic sequence, i.e., have $a$ common difference or the second difference is a constant.

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Example:

From the second example of a quadratic sequence above

The second difference is a constant of 4, therefore showing that the sequence given is an example of a quadratic sequence. It can be shown that $a$ is given by $a=\frac{\mathrm{second\; difference}}{2}$.

There are three methods, listed below, for finding the other two characteristic values $b$ and $c$:

1. use properties of the quadratic sequence alone with how to find $a$ given above to determine $b$ and $c$ ;
2. memorise a set of simultaneous equations to find $a$, $b$ and $c$ ;
3. use a method for determining the values of $a$, $b$ and $c$ by building on the work already completed.

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Videos and Questions:

Quadratic nth term – Version 1 Video:

Quadratic nth term – Version 2 Video:

Quadratic nth term – Version 3 Video:

Practice Questions;

Textbook Exercise.

(Care of corbettmaths.com)

The Nth Term of a Quadratic Sequence:

Exam Questions;

Quadratic Sequences;

Solutions.

(Care of www.mathsgenie.co.uk)

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Quadratic Sequence n^th Term

Prerequisite material:

Quadratics Sequences

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The following is a method for finding the n^th term of a quadratic sequence, $U_n$

1. Confirm the second difference is constant, then, $$U_n=a{n}^2+bn+c$$
2. To find $c$ use the zeroth term, $$U_0=a\times 0^2+b\times 0+c$$

   $$U_0=c$$

3. To find $a$ and $b$ by using the first terms and second term: $$U_1=a\times 1^2+b\times 1+c$$

   $$U_1=a+b+c$$

   $$U_2=a\times 2^2+b\times 2+c$$

   $$U_2=4a+2b+c$$

4. Solve this pair of simultaneous to find $a$ and $b$. You may know from other study how to find $a$ using

   $$a=\frac{\text{second difference}}{2}$$

5. Then

   $$U_n=a{n}^2+bn+c$$

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Quadratic Sequence n^th Term

We can continue the work which has already been carried out.

Using the calculated value of $a[=\frac{\text{second difference}}{2}]$

We can then calculate the difference between $U_n$ and $a{n}^2$ as we know this is equal to $bn+c$. So,

This new sequence is an arthmetic. So we can determine it's $n^{\text{th}}$ term as $-3\mathrm{n<mo>+</mo><mn>4</mn>}$.

Therefore, the orginal quadratic sequence has an $n^{\text{th}}$ term expression of

$$U_n=2n^2-3\mathrm{n<mo>+</mo><mn>4</mn>}$$

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Questions:

Find the next two terms and the nth term expression for each of these quadratic sequences.

1. 1, 5, 11, 19, 29, ...

2. 9, 20, 33, 48, 65, ...

3. 4, 10, 14, 16, 16, ...

4. 4, 11, 20, 31, 44, ...

5. 0, 1, 6, 15, 28, ...

6. 10, 18, 28, 40, 54, ...

7. 5, 9, 17, 29, 45, ...

8. -5, -3, 3, 13, 27, ...

9. 5, 1, -5, -13, -23, ...

10. -4, 0, 2, 2, 0, ...

Solutions:

1. 41 and 55, $U_n=n^2+n-1$

2. 85 and 105, $U_n=n^2+8n$

3. 14 and 10, $U_n={\mathrm{-n}}^2+8n-4$

4. 59 and 76, $U_n=n^2+4n-1$

5. 45 and 66, $U_n=2n^2-5n+3$

6. 70 and 88, $U_n=n^2+5n+4$

7. 65 and 89, $U_n=2n^2-2n+5$

8. 43 and 61, $U_n=2n^2-4n-3$

9. and , $U_n=n^2+n-1$

10. and , $U_n=n^2+n-1$

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Further Reading:

Cubic Sequences
(Care of www.geeksforgeeks.org)

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Proof

The general form of a quadratic sequence is given by

$$U_n=a{n}^2+bn+c$$

The first term is given by

$$U_1=a+b+c$$

The difference between each term is

$$U_{n+1}-U_n=a{(n+1)}^2+b(n+1)+c-(a{n}^2+bn+c)$$ $$=a(n^2+2n+1)+bn+b+c-a{n}^2-bn-c$$ $$=a{n}^2+2an+a+b-a{n}^2$$ $$=2an+a+b$$

So the difference between terms is a quadratic sequence is

$$U_{n+1}-U_n=2an+a+b$$

& the first difference between terms is

$$U_2-U_1=3a+b$$

Now taking the difference of the differences between terms, or, sometimes called, the second difference, that is, a difference minus the previous difference,

$$(U_{n+2}-U_{n+1})-(U_{n+1}-U_n)$$ $$=(2a(n+1)+a+b)-(2an+a+b)$$ $$=2a$$

So the difference between the differences in terms (or the second difference) is the constant $2a$.

We can use these results to help determine the values of $a$, $b$ & $c$.

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