`a_1 = 0, a_2 = 8, a_4 = 30` Find a quadratic model for the sequence with the indicated terms.

Monday, March 9, 2015

$\displaystyle{a}_{{1}}={0},{a}_{{2}}={8},{a}_{{4}}={30}$ Find a quadratic model for the sequence with the indicated terms.

The given sequence is:

$\displaystyle{a}_{{1}}={0}$ , $\displaystyle{a}_{{2}}={8}$ , $\displaystyle{a}_{{4}}={30}$

To determine its quadratic model, apply the formula

$\displaystyle{f{{\left({n}\right)}}}={a}{{n}}^{{2}}+{b}{n}+{c}$

where f(n) represents the nth term of the sequence, $\displaystyle{f{{\left({n}\right)}}}={a}_{{n}}$ .

So, plug-in the first term of the sequence.

$\displaystyle{0}={a}{{\left({1}\right)}}^{{2}}+{b}{\left({1}\right)}+{c}$

$\displaystyle{0}={a}+{b}+{c}$ (Let this be EQ1.)

Plug-in too the second term of the sequence.

$\displaystyle{8}={a}{{\left({2}\right)}}^{{2}}+{b}{\left({2}\right)}+{c}$

$\displaystyle{8}={4}{a}+{2}{b}+{c}$ ...

The given sequence is:

$\displaystyle{a}_{{1}}={0}$ , $\displaystyle{a}_{{2}}={8}$ , $\displaystyle{a}_{{4}}={30}$

To determine its quadratic model, apply the formula

$\displaystyle{f{{\left({n}\right)}}}={a}{{n}}^{{2}}+{b}{n}+{c}$

where f(n) represents the nth term of the sequence, $\displaystyle{f{{\left({n}\right)}}}={a}_{{n}}$ .

So, plug-in the first term of the sequence.

$\displaystyle{0}={a}{{\left({1}\right)}}^{{2}}+{b}{\left({1}\right)}+{c}$

$\displaystyle{0}={a}+{b}+{c}$ (Let this be EQ1.)

Plug-in too the second term of the sequence.

$\displaystyle{8}={a}{{\left({2}\right)}}^{{2}}+{b}{\left({2}\right)}+{c}$

$\displaystyle{8}={4}{a}+{2}{b}+{c}$ (Let this be EQ2.)

And, plug-in the 4th term of the sequence.

$\displaystyle{30}={a}{{\left({4}\right)}}^{{2}}+{b}{\left({4}\right)}+{c}$

$\displaystyle{30}={16}{a}+{4}{b}+{c}$ (Let this be EQ3.)

To solve for the values of a, b and c, apply elimination method of system of equations. In this method, a variable or variables should be removed.

Let's eliminate c. To do so, subtract EQ1 from EQ2.

EQ2: $\displaystyle{8}={4}{a}+{2}{b}+{c}$

EQ1: $\displaystyle-{\left({0}={a}+{b}+{c}\right)}$

$\displaystyle----------------$

$\displaystyle{8}={3}{a}+{b}$ (Let this be EQ4.)

Let's eliminate c again. This time, subtract EQ2 from EQ3.

EQ3: $\displaystyle{30}={16}{a}+{4}{b}+{c}$

EQ2: $\displaystyle-{\left({8}={4}{a}+{2}{b}+{c}\right)}$

$\displaystyle----------------$

$\displaystyle{22}={12}{a}+{2}{b}$

And this simplifies to:

$\displaystyle\frac{{22}}{{2}}=\frac{{{12}{a}+{2}{b}}}{{2}}$

$\displaystyle{11}={6}{a}+{b}$ (Let this be EQ5.)

Then, eliminate b. To do so, subtract EQ4 from EQ5.

EQ5: $\displaystyle{11}={6}{a}+{b}$

EQ4: $\displaystyle-{\left({8}={3}{a}+{b}\right)}$

$\displaystyle--------------$

$\displaystyle{3}={3}{a}$

Isolating the a, it becomes:

$\displaystyle\frac{{3}}{{3}}=\frac{{{3}{a}}}{{3}}$

$\displaystyle{1}={a}$

Then, plug-in the value of a to either EQ4 or EQ5. Let's use EQ4.

$\displaystyle{8}={3}{a}+{b}$

$\displaystyle{8}={3}{\left({1}\right)}+{b}$

$\displaystyle{8}={3}+{b}$

$\displaystyle{8}-{3}={3}-{3}+{b}$

$\displaystyle{5}={b}$

And, plug-in the values of a and b to either EQ1, EQ2 or EQ3. Let's use EQ1.

$\displaystyle{0}={a}+{b}+{c}$

$\displaystyle{0}={1}+{5}+{c}$

$\displaystyle{0}={6}+{c}$

$\displaystyle{0}-{6}={6}-{6}+{c}$

$\displaystyle-{6}={c}$

Now that the values of a, b and c are known, plug-in them to:

$\displaystyle{f{{\left({n}\right)}}}={a}{{n}}^{{2}}+{b}{n}+{c}$

$\displaystyle{f{{\left({n}\right)}}}={\left({1}\right)}{{n}}^{{2}}+{5}{n}+{\left(-{6}\right)}$

$\displaystyle{f{{\left({n}\right)}}}={{n}}^{{2}}+{5}{n}-{6}$

Replacing the f(n) with an, it becomes:

$\displaystyle{a}_{{n}}={{n}}^{{2}}+{5}{n}-{6}$

Therefore, the quadratic model of the sequence is $\displaystyle{a}_{{n}}={{n}}^{{2}}+{5}{n}-{6}$ .

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