Nice way. I always solved it like i learned in school a2-a1=b1 a3-a2=b2..an-a(n-1)=b(n-1) so summing all we get an=a1 +(b1+b2..+b(n-1)) which is the sum ofn-1 arithemetic series.
Very nice video and I've loved the trick with A! I think that you could use one more trick for the first difference in order to get B, 'cuz the formula for the first difference would be A(2n+1) + B. Taking the last example, once we know that A = 6, taking the first term of the first difference we get 18 = 6(2*(1) + 1) + B = 18 + B, and then B = 0. Finally, C can be obtained with the first term of the sequence, using that A(1)² + B(1) + C = 1, we have 6 + 0 + C = 1 and therefore C = -5.
You can create a new series by subtracting An^2 from the original series (you only need the first two values in this new series). Then the first difference will be equal to B. Then, subtracting A1 and B1 from the first value in the first sequence gives you C
wala na 'kong magets. kaso pag bumagsak hindi ako makakapagtapos, pag 'di ako nakapagtapos hindi ako magkakatrabaho, pag wala akong trabaho magkakasakit ako, kapag nagkasakit ako mamatay ako kasi nga wala akong pambili ng gamot kasi wala akong trabaho. kaya shet panoodin ko ulit 2hs loop (nahihilo, nalilito)
A shortcut you can use is to generate a_0 (because you can calculate the difference between a_0 and a_1), and then you get C for free because: a_0=A*0^2+B*0+C=C.
