Notes: `a_0 = -3, a_2 = 1, a_4 = 9` Find a quadratic model for the sequence with the indicated terms.

Saturday, September 3, 2011

You need to remember what a quadratic model is, such that:

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

The problem provides the following information, such that:

$\displaystyle{a}_{{0}}=-{3}\Rightarrow{f{{\left({0}\right)}}}={a}\cdot{{0}}^{{2}}+{b}\cdot{0}+{c}\Rightarrow{c}=-{3}$

$\displaystyle{a}_{{2}}={1}\Rightarrow{f{{\left({2}\right)}}}={a}\cdot{{2}}^{{2}}+{b}\cdot{2}+{c}\Rightarrow{4}{a}+{2}{b}+{c}={1}$

$\displaystyle{a}_{{4}}={9}\Rightarrow{f{{\left({4}\right)}}}={a}\cdot{{4}}^{{2}}+{b}\cdot{4}+{c}\Rightarrow{16}{a}+{4}{b}+{c}={9}$

You need to replace -3 for c in equation $\displaystyle{4}{a}+{2}{b}+{c}={1}:$

$\displaystyle{4}{a}+{2}{b}-{3}={1}\Rightarrow{4}{a}+{2}{b}={4}\Rightarrow{2}{a}+{b}={2}$

You need to replace -3 for c in equation $\displaystyle{16}{a}+{4}{b}+{c}={9}$ :

$\displaystyle{16}{a}+{4}{b}-{3}={9}\Rightarrow{16}{a}+{4}{b}={12}\Rightarrow{4}{a}+{b}={3}$

Subtract $\displaystyle{2}{a}+{b}={2}$ from $\displaystyle{4}{a}+{b}={3},$ such that:

$\displaystyle{4}{a}+{b}-{2}{a}-{b}={3}-{2}$

$\displaystyle{2}{a}={1}\Rightarrow{a}=\frac{{1}}{{2}}$

Replace $\displaystyle\frac{{1}}{{2}}$ for a in equation $\displaystyle{2}{a}+{b}={2}$ such that:

$\displaystyle{2}\cdot{\left(\frac{{1}}{{2}}\right)}+{b}={2}\Rightarrow{1}+{b}={2}\Rightarrow{b}={1}$

Hence, the quadratic model for the given sequence is $\displaystyle{a}_{{n}}={\left(\frac{{1}}{{2}}\right)}{{n}}^{{2}}+{n}-{3}.$

Notes