The sequence is solution of the 'difference equation'...
(1)
... which can be written as...
(2)
The 'standard solution' of (1) is...
(3)
Regarding the solving procedure of (1) see the tutorial post on the difference equation in the section 'Discrete Mathematics'...
Kind regards
What you have there are the first differences, when you take the differences again you get a constant, and you should know that the constant n-th differences mean the original sequence was a polynomial of degree n in the index. Hence as the second differences are constant the original sequence is a quadratic in the index.
CB
Let , then:
from the first term in the sequence, and:
from the 2nd term, and:
from the third.
These are all the terms that need be used since the three simultaneous equations will have a unique solution (and it will be the same for three equations formed from any three distinct terms in the sequence).
Also Newton has a systematic way of doing this from the difference table that you may have covered in class.
CB
It is a question of 'initial index'... if the initial index is 0, then the difference equation is...
(1)
... and its solution is...
(2)
... if the initial index is 1, then the difference equation is...
(3)
... and its solution is...
(4)
In any case You have as solution the sequence 1,4,8,13,19,...
Kind regards