Given the quadratic number pattern: \( -5 ; -4 ; -1 ; 4 ; \ldots \) 311 Determine the \( n^{*} \) term of the quadratic number patterns in the form \( T_{n}=a n^{2}+b n+c \).
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To find the \( n^{*} \) term of the quadratic number pattern, we start by observing the differences between the consecutive terms: - The first differences: \( -4 - (-5) = 1 \), \( -1 - (-4) = 3 \), \( 4 - (-1) = 5 \) → \( 1, 3, 5 \) - The second differences: \( 3 - 1 = 2 \), \( 5 - 3 = 2 \) → consistent second difference of 2. This suggests that the sequence is quadratic, and we can represent it as: \[ T_n = an^2 + bn + c \] Since the second difference is 2, we find \( a = 1 \) (as the second difference is \( 2a \)). Now our equation is: \[ T_n = n^2 + bn + c \] Using the given terms: - For \( n=1 \), \( T_1 = -5 \) → \( 1 + b + c = -5 \) → \( b + c = -6 \) - For \( n=2 \), \( T_2 = -4 \) → \( 4 + 2b + c = -4 \) → \( 2b + c = -8 \) - For \( n=3 \), \( T_3 = -1 \) → \( 9 + 3b + c = -1 \) → \( 3b + c = -10 \) We can solve these equations in sequence. From \( b + c = -6 \) and \( 2b + c = -8 \), we subtract the first from the second: \[ (2b + c) - (b + c) = -8 + 6 \] Thus: \[ b = -2. \] Substitute \( b = -2 \) into \( b + c = -6\): \[ -2 + c = -6 \] So: \[ c = -4. \] Now, our equation becomes: \[ T_n = n^2 - 2n - 4. \] To determine the \( n^{*} \) term, we substitute \( n = 311 \): \[ T_{311} = 311^2 - 2(311) - 4 = 96721 - 622 - 4 = 96195. \] Therefore, the \( n^{*} \) term is \( 96195. \)