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=15+97.99.101.1038.

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=3+97.99.1016 .

Vì vậy A=M−2N=15+97.99.101.1038−2.3+97.99.1016

=15+97.99.101.1038−1+97.33.101

=15+97.99.101.103−8−8.97.33.1018

=7+97.33.101.3.103−88

=7+97.33.101.3018

= 12 164 201.

Vậy A = 12 164 201.

Tìm số tự nhiên x, biết: x chia hết cho 9 và x < 40.