Answer:
y=n^2+3n+4
Step-by-step explanation:
Sequence,S: 8,14,22,32 ,44, 58, 74...
First differences of S: 6,8,10,12,14,16,...
Second differences of S: 2,2,2,2,2,2,....
It is indeed a quadratic since the 1st differences that are the same are the second differences.
So the 1st term is 8. If the 0th term was listes it would be 4 since 8-4=4 and 6-4=2.
This means our quadratic it is in the form
y=an^2+bn+c where c=4.
So we have y=an^2+bn+4.
Now we can find a and b using two points from our sequence to form a system of equations to solve.
(1,8) gives us 8=a(1)^2+b(1)+4
Simplifying gives 8=a+b+4
Subtracting 4 on both sides gives: 4=a+b
(2,14) gives us 14=a(2)^2+b(2)+4
Simplifying gives 14=4a+2b+4
Subtracting 4 on both sides gives 10=4a+2b
So we have the following system to solve:
4=a+b
10=4a+2b
I'm going to solve using elimination/linear combination style.
Dividing second equation by 2 gives the system as:
4=a+b
5=2a+b
Subtract the 2nd equation from the first gives:
-1=-1a
This implies a=1 since -1=-1(1) is true.
If a+b=4 and a=1, then b=3. Because 3+1=4.
So the equation is y=1n^2+3n+4 or y=n^2+3n+4.
