Nth term Quadratic Sequences?

Question

Hey. I'm rather stuck on this part. I'm perfectly fine on linear squences but qudratic is challenging.

For example,

1. Find the nth term of this sequence: 2, 5, 9, 11 ...

For this i used the changing equation type formula which is a+(n-1)d + 1/2 (n-1) )(n-2)C. So i then add in the values which are 2 + (n-1)3 + 1/2 (n-1) (n-2) x 1. Then i break it down by doing 2 + 3n - 3 + 1/2n^2 - 1 1/2n + 1.

If i leave it like this in an examination is it ok? Or must i simplify it? Can i not leave it like that?

All comments are much appreciated. Many thanks!

Answer 1

Is the 11 correct?  It would be more orderly if the sequence were 2, 5, 9, 14, ...

Answer 2

It's just an arithmetic series isn't it.

The answer usually looks like (n+a)(n+b)/2

Plug in n=1 and n=2 to find a and b.

If it's squares in the series, answer is (n+a)(n+b)(n+c)*constant..