Ta có
$S = 2\times 4 + 4 \times 6 + \cdots + 1998 \times 2000$
$= 2 \times 1 \times 2 \times 2 + 2 \times 2 \times 3 \times 2 + \cdots + 2 \times 999 \times 1000 \times 2$
$= 4(1 \times 2 + 2 \times 3 + \cdots + 999 \times 1000)$
Đặt
$A = 1 \times 2 + 2 \times 3 + \cdots + 999 \times 1000$
Nhân $3$ vào 2 vế ta có
$3A = 1 \times 2 \times 3 + 2 \times 3 \times 3 + \cdots + 999 \times 1000 \times 3$
$= 1 \times 2 \times 3 + 2 \times 3 \times (4-1) + 3 \times 4 \times (5-2) + \cdots + 999 \times 1000 \times (1001 - 998)$
$= (1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots + 999 \times 1000 \times 1001) - (1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots + 998 \times 999 \times 1000)$
$= 999 \times 1000 \times 1001$
Suy ra
$A = 333 \times 1000 \times 1001$
Vậy ta có
$S = 4A = 333 \times 4000 \times 1001$
