The sequence

$\text{If you look carefully at the sequence, you'll get a general formula : } \\ n^{th} = 2n(n+1) \text{ ... } \fbox{1}\\ \text{So, }-n^{th}\text{ term would be} \\ -n^{th} = 2(-n)(-n + 1) = -2n ( 1-n) = 2n(n-1) \\ \text{Therefore, }-(n+1)^{th} \text{ term would be :}\\ -(n+1)^{th} = 2(n+1)((n+1)-1) = 2(n+1)(n) = 2n(n+1) \text{ ... }\fbox{2} \\ \text{Now, on subtracting } \fbox{1} \text{ and } \fbox{2} \text{ you get :}\\ n^{th} - (-(n+1)^{th}) = 2n(n+1) - (2n(n+1)) \\ \implies n^{th} - (-(n+1)^{th}) = 0 \\ \text{Therefore, the answer is }\fbox{0} \\ \hrule \\ \text{Now to find the sequence.} \\ \text{Let the general formula be f(x). Given that f(x) have 2 zeros, therefore f(x) is a quadratic equation.} \\ \text{Roots are 0, -1} \\ \implies x = 0\div a, x = -1 \\ \implies f(x) = (ax)(x+1) \\ \text{Therefore, on substituting values for "x" and matching it with sequence : } \\ \text{f(1) - f(0) = 4} \\ \implies 1a(1+1) - 0a(0+1) = 4\\\implies a(2) + 0 = 4\\\implies 2a = 4\\\implies \fbox{a = 2} \\ \text{Therefore the formula is } f(x) = 2n(n+1)$