Question

[Eromanga Sensei]

Tính:

P = 1.2²+2.3²+3.4²+....+99.100²

^{HELP ME, chìu mình đi học rùi T_T}

Answer 1

[Nàng tiên cá]

Đồng cảnh ngộ.Huhu

Answer 2

[»βέ•Ҫɦαηɦ«]

Ta có : P = 1.2.2 + 2.3.3 + ....+ 99.100.100

=1.2.(3 - 1) + 2.3.(4 - 1) + ....+99.100.(101 - 1)

= (1.2.3 + 2.3.4 + .... + 99.100.101) - (2.3 + 3.4+.....+99.100)

Đặt B = 1.2.3 + 2.3.4 + 4.5.6 +...+ 99.100.101

4B = 1.2.3.(4 - 0)+2.3.4.(5 - 1) + ... + (99.100.101(102 - 98)

4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 +...+ 99.100.101.102 - 98.99.100.101

4B = 99.100.101.102

4B = 101989800

  B = 25497450

Đặt C = 1.2 + 2.3 + 3.4 +...+ 99.100

3C = 1.2.(3 - 0) + 2.3.(4 - 1) +...+ 99.100.(101 - 98)

3C = 1.2.3 + 2.3.4 - 1.2.3 +...+ 99.100.101 - 98.99.100

3C = 99.100.101

3C = 999900

  C = 999900 : 3

  C = 333300

Vậy: A = 25497450 – 333300 = 25164150