So, the nth term formula will take the form an^2+bn+c where a, b, and c are numbers to be determined. Firstly, we have to find the differences between the terms in the sequences, and then find the difference between the differences. Doing so, we get
The second difference are the same, as expected, therefore a is half of the second difference and so a=2.
Now we need to find b and c. To do so, we need to first write out the terms in our sequence again. Then, we need to take the part of the nth term formula that we now know, which is 2n^2, substitute the values of n from 1 to 5 into this, and write the results below the original sequence. Finally, we will subtract each member of the second row from the member of the first row that sits above it and write the results in a third row. This should look like:
Then, d_n is a linear sequence whose nth term formula is precisely bn+c, so finding it will give us the missing parts of our quadratic nth term formula. So, looking at the differences between the terms of d_n, we see
The difference is -5, so the nth term must be -5n+c. Then, for n=1, we get -5n=-5, whereas the first term is not -5, but -2. To get from -5 to -2 we have to add 3, so we must have that c=3, and thus the nth term for d_n is
Therefore, combining this with the first time in the quadratic that we found earlier, we get the nth term formula of the quadratic to be