In a quadratic rumber puttern, it is given that the seeend constont difference is -6 , the first term is 1 and the fifth term is -55 . Oetermine the geveral formula in this form \( h=a n^{2}+b n+c \).

To derive the general formula \( h = an^2 + bn + c \), we know that the second constant difference is -6, which tells us that \( a = -3 \) (since the second difference is equal to \( 2a \)). So we have \( h = -3n^2 + bn + c \). We can use the given terms to find \( b \) and \( c \). Given that the first term \( h(1) = 1 \), we can set up the equation: \[ -3(1)^2 + b(1) + c = 1 \implies -3 + b + c = 1 \implies b + c = 4. \] Next, for the fifth term \( h(5) = -55 \), we have: \[ -3(5)^2 + b(5) + c = -55 \implies -75 + 5b + c = -55 \implies 5b + c = 20. \] Now we have a system of equations: 1. \( b + c = 4 \) 2. \( 5b + c = 20 \) Subtracting the first from the second gives: \[ 4b = 16 \implies b = 4. \] Substituting \( b = 4 \) back into the first equation gives: \[ 4 + c = 4 \implies c = 0. \] Thus, the general formula we are looking for is: **\( h = -3n^2 + 4n \)**. This quadratic formula encapsulates the upward mountain we can climb through our number pattern!