Do $(u_n)$ là csc nên
$u_n = (n-1)d + u_1$
Do $(v_n)$ là csn nên
$v_n = v_1 . q^{n-1}$
Vậy ta có
$S = u_1 v_1 + (u_1 + d).v_1.q + \cdots + [u_1 + (n-1)d] . v_1 . q^{n-1}$
$= u_1 v_1 + (u_1 v_1 q + dv_1q) + \cdots + [u_1 v_1 q^{n-1} + v_1 q^{n-1} (n-1)d]$
$= u_1 v_1(1 + q + \cdots + q^{n-1})+ v_1(dq + \cdots + (n-1)dq^{n-1})$
$= u_1 v_1 . \dfrac{1-q^n}{1-q} + v_1(dq + \cdots + (n-1)dq^{n-1})$