What is the nth term rule of the quadratic sequence below?
8,14,22,32 ,44, 58, 74...

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.