Find the first term, the common difference, and tn for a series having Sn = 4n^2+n.

Find the first term, the common difference, and tn for a series having Sn = 4n^2+n.

$\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n^2+n \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n^2+n - [ 4(n-1)^2+(n-1) ] \\ t_n &=&4n^2+n - 4(n-1)^2 - (n-1) ] \\ t_n &=&4n^2+n - 4(n-1)^2 - n +1 ] \\ t_n &=&4n^2 - 4(n-1)^2 +1 ] \\ t_n &=&4n^2 - 4(n^2-2n+1) +1 ] \\ t_n &=&4n^2 - 4n^2 +8n-4 +1 ] \\ t_n &=&8n-4 +1 ] \\ \mathbf{t_n} & \mathbf{=} & \mathbf{8n-3} \\\\ t_1 &=& 8\cdot 1 - 3 \\ t_1 &=& 8-3\\ \mathbf{t_1} &\mathbf{=}& \mathbf{5}\\\\ d &=& t_n - t_{n-1} \\ d &=& 8n-3 -[ 8\cdot (n-1)-3 ]\\ d &=& 8n-3 - 8\cdot (n-1) +3 \\ d &=& 8n - 8\cdot (n-1) \\ d &=& 8n - 8n+8 \\ \mathbf{d} &\mathbf{=}& \mathbf{8} \\ \end{array}$