Notes: `a_0 = 3, a_2 = 0, a_6 = 36` Find a quadratic model for the sequence with the indicated terms.

Wednesday, October 21, 2015

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:

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

$\displaystyle{a}_{{2}}={0}\Rightarrow{f{{\left({2}\right)}}}={a}\cdot{{2}}^{{2}}+{b}\cdot{2}+{c}\Rightarrow{4}{a}+{2}{b}+{c}={0}$

$\displaystyle{a}_{{6}}={36}\Rightarrow{f{{\left({6}\right)}}}={a}\cdot{{6}}^{{2}}+{b}\cdot{6}+{c}\Rightarrow{36}{a}+{6}{b}+{c}={36}$

You need to replace 3 for c in the next two equations, such that:

$\displaystyle{4}{a}+{2}{b}+{3}={0}\Rightarrow{4}{a}+{2}{b}=-{3}$

$\displaystyle{36}{a}+{6}{b}+{3}={36}\Rightarrow{36}{a}+{6}{b}={33}\Rightarrow{18}{a}+{2}{b}={11}$

Subtract the equation $\displaystyle{4}{a}+{2}{b}=-{3}$ from the equation $\displaystyle{18}{a}+{2}{b}={11}$ :

$\displaystyle{18}{a}+{2}{b}-{4}{a}-{2}{b}={11}+{3}$

$\displaystyle{14}{a}={14}\Rightarrow{a}={1}$

Replace 1 for a in equation $\displaystyle{4}{a}+{2}{b}=-{3}$ , such that:

$\displaystyle{4}+{2}{b}=-{3}\Rightarrow{2}{b}=-{7}\Rightarrow{b}=-\frac{{7}}{{2}}$

Hence, the quadratic model for the given sequence is $\displaystyle{a}_{{n}}={{n}}^{{2}}-{\left(\frac{{7}}{{2}}\right)}\cdot{n}+{3}.$

Notes