The given sequence is:
`a_1 = 0` , `a_2=8` , `a_4=30`
To determine its quadratic model, apply the formula
`f(n) = an^2 + bn + c`
where f(n) represents the nth term of the sequence, `f(n)=a_n` .
So, plug-in the first term of the sequence.
`0=a(1)^2 + b(1) + c`
`0=a+b+c ` (Let this be EQ1.)
Plug-in too the second term of the sequence.
`8=a(2)^2+b(2)+c`
`8=4a+2b+c` (Let this be EQ2.)
And, plug-in the 4th term of the sequence.
`30=a(4)^2+b(4)+c`
`30=16a+4b+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: `8=4a+2b+c`
EQ1: `-(0=a+b+c)`
`----------------`
`8=3a+b` (Let this be EQ4.)
Let's eliminate c again. This time, subtract EQ2 from EQ3.
EQ3: `30=16a+4b+c`
EQ2: `-(8=4a+2b+c)`
`----------------`
`22=12a+2b`
And this simplifies to:
`22/2=(12a+2b)/2`
`11=6a+b ` (Let this be EQ5.)
Then, eliminate b. To do so, subtract EQ4 from EQ5.
EQ5: `11=6a+b`
EQ4: `-(8=3a+b)`
`--------------`
`3=3a`
Isolating the a, it becomes:
`3/3=(3a)/3`
`1=a`
Then, plug-in the value of a to either EQ4 or EQ5. Let's use EQ4.
`8=3a+b`
`8=3(1) + b`
`8=3+b`
`8-3=3-3+b`
`5=b`
And, plug-in the values of a and b to either EQ1, EQ2 or EQ3. Let's use EQ1.
`0=a+b+c`
`0=1+5+c`
`0=6+c`
`0-6=6-6+c`
`-6=c`
Now that the values of a, b and c are known, plug-in them to:
`f(n)=an^2+bn+c`
`f(n)=(1)n^2+5n+(-6)`
`f(n)=n^2+5n-6`
Replacing the f(n) with an, it becomes:
`a_n=n^2+5n-6`
Therefore, the quadratic model of the sequence is `a_n=n^2+5n-6` .
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