Notes: `-2, 1, 6, 13, 22, 33.....` Decide whether the sequence can be represented perfectly by a linear or a quadratic model. If so, then find the...

Wednesday, August 5, 2015

$\displaystyle-{2},{1},{6},{13},{22},{33}\ldots..$ Decide whether the sequence can be represented perfectly by a linear or a quadratic model. If so, then find the...

Firstly we need to determine whether the series is linear or quadratic. A linear sequence is a sequence of numbers in which there is a first difference between any consecutive terms is constant. However, a quadratic sequence is a sequence of numbers in which there is a second difference between any consecutive terms is constant.

Let's determine if the above sequence has a first difference:

$\displaystyle{x}_{{1}}={T}_{{2}}-{T}_{{1}}={1}-{\left(-{2}\right)}={3}$

$\displaystyle{x}_{{2}}={T}_{{3}}-{T}_{{2}}={6}-{1}={5}$

$\displaystyle{x}_{{3}}={T}_{{4}}-{T}_{{3}}={13}-{6}={7}$

From above we observe there is no first difference, now let's determine the second difference:

$\displaystyle{x}_{{2}}-{x}_{{1}}={5}-{3}={2}$

$\displaystyle{x}_{{3}}-{x}_{{2}}={7}-{5}={2}$

From above we observe that the second difference is constant. Therefore the sequence is perfectly quadratic.

Let's determine the quadratic model using the following formula:

$\displaystyle{T}_{{n}}={a}{{n}}^{{2}}+{b}{n}+{c}$

where: T_n = term value, ,n = term, variables: a,b,c

$\displaystyle{2}{a}$ = second difference

$\displaystyle{2}{a}={2}$

$\displaystyle{a}={1}$

Let's find variable b:

$\displaystyle{3}{a}+{b}=$ first difference between term 2 and term 1

$\displaystyle{3}{\left({1}\right)}+{b}={3}$ (substitute for a and first difference between term 2 and term1)

$\displaystyle{b}={3}-{3}={0}$

Lastly we are finding variable c:

$\displaystyle{a}+{b}+{c}=$ value of term 1

$\displaystyle{1}+{0}+{c}=-{2}$ (substitute for a, b and value of term 1

$\displaystyle{c}=-{2}-{1}=-{3}$

Now we have determined the variables, we can develop our model:

$\displaystyle{T}_{{n}}={1}{{\left({n}\right)}}^{{2}}+{0}{\left({n}\right)}-{3}={{n}}^{{2}}-{3}$

Let's double check our formula using term 4:

$\displaystyle{T}_{{4}}={{\left({4}\right)}}^{{2}}-{3}={16}-{3}={13}$

SUMMARY:

Sequence: Quadratic

Model: $\displaystyle{T}_{{n}}={{n}}^{{2}}-{3}$

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