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

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1. 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.

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