Tính: 1.3² + 3.5² + 5.7² + ... + 97.99².

Đặt A = 1.3² + 3.5² + 5.7² + ... + 97.99².

= 1.3.3 + 3.5.5 + 5.7.7 + ... + 97.99.99.

= 1.3.(5 – 2) + 3.5.(7 – 2) + 5.7.(9 – 2) + ... + 97.99.(101 – 2).

= (1.3.5 – 1.3.2) + (3.5.7 – 3.5.2) + (5.7.9 – 5.7.2) + ... + (97.99.101 – 97.99.2).

= (1.3.5 + 3.5.7 + 5.7.9 + ... + 97.99.101) – 2.(1.3 + 3.5 + 5.7 + ... + 97.99).

Đặt M = 1.3.5 + 3.5.7 + 5.7.9 + ... + 97.99.101 và N = 1.3 + 3.5 + 5.7 + ... + 97.99.

Ta có 8M = 1.3.5.8 + 3.5.7.(9 – 1) + 5.7.9.(11 – 3) + ... + 97.99.101.(103 – 95).

= 1.3.5.8 + (3.5.7.9 – 1.3.5.7) + (5.7.9.11 – 3.5.7.9) + ... + (97.99.101.103 – 95.97.99.101.

= 1.3.5.8 – 1.3.5.7 + 97.99.101.103

= 1.3.5.(8 – 7) + 97.99.101.103

= 15 + 97.99.101.103.

Suy ra $M=\frac{15+\mathrm{97.99.101.103}}{8}$ .

Lại có 6N = 1.3.6 + 3.5.6 + 5.7.6 + ... + 97.99.6

= 1.3.(5 + 1) + 3.5.(7 – 1) + 5.7.(9 – 3) + ... + 97.99.(101 – 95)

= 1.3.5 + 1.3.1 + 3.5.7 – 3.5.1 + 5.7.9 – 5.7.3 + ... + 97.99.101 – 97.99.95

= 1.3.1 + 97.99.101

= 3 + 97.99.101.

Suy ra $N=\frac{3+\mathrm{97.99.101}}{6}$  .

Vì vậy $A=M-2N=\frac{15+\mathrm{97.99.101.103}}{8}-2.\frac{3+\mathrm{97.99.101}}{6}$

$=\frac{15+\mathrm{97.99.101.103}}{8}-\left(1+\mathrm{97.33.101}\right)$

$=\frac{15+\mathrm{97.99.101.103}-8-\mathrm{8.97.33.101}}{8}$

$=\frac{7+\mathrm{97.33.101.}\left(3.103-8\right)}{8}$

$=\frac{7+\mathrm{97.33.101.301}}{8}$

= 12 164 201.

Vậy A = 12 164 201.