I came across a problem which follows the series: $2,3,6,11,18$ and so on..
It can be observed that the common difference of the series was in an arithmetic progression..
Using the regular method of supposing the general term as $an^2+bn+c$ and plugging in the values $n=1,2,3$ for the first, second and third term of the series respectively, I got the equation $n^2-2n+3$..
For $n=0$, the zeroth term (which doesn't exist for a definite sequence) would give the value 3 from the above derived equation...
If the common difference (AP series) is extended backwards, the previous number in the indefinite series of the above given series will be 3 which basically is the zeroth term, and the previous common difference being $-1...$
Is this a special case or can I always extend the series backwards and put $n=0$ for a series for a faster calculation of c thereby giving a faster calculation of the equation of the first term instead of solving 3 equations in 3 variables?