Consider this sequence series: n^2+n+1, between which two consecutive terms in the sequence is the first difference equal to 2?

To find out between which two consecutive terms the first difference equals 2 in the sequence defined by \( a_n = n^2 + n + 1 \), we first need to calculate the first difference \( a_{n+1} - a_n \). The sequence's elements are given by: \[ a_n = n^2 + n + 1 \] \[ a_{n+1} = (n+1)^2 + (n+1) + 1 = n^2 + 2n + 1 + n + 1 + 1 = n^2 + 3n + 3 \] Now, we find the first difference: \[ a_{n+1} - a_n = (n^2 + 3n + 3) - (n^2 + n + 1) \] \[ = n^2 + 3n + 3 - n^2 - n - 1 \] \[ = 3n + 3 - n - 1 \] \[ = 2n + 2 \] We need to find the smallest \( n \) such that the first difference \( 2n + 2 = 2 \). Solving the equation: \[ 2n + 2 = 2 \] Subtracting 2 from both sides: \[ 2n = 0 \] Dividing both sides by 2: \[ n = 0 \] Now, we evaluate the two consecutive terms at \( n = 0 \): \[ a_0 = 0^2 + 0 + 1 = 1 \] \[ a_1 = 1^2 + 1 + 1 = 3 \] Thus, the first difference \( a_1 - a_0 = 3 - 1 = 2 \). Therefore, the first difference is equal to 2 between the consecutive terms \( a_0 \) and \( a_1 \). The answer is that the first difference equal to 2 occurs between the terms \( a_0 \) and \( a_1 \).