a/
S=1.2.(3-1)+2.3.(4-1)+3.4.(5-1)+...+99.100.(101-1)=
=1.2.3+2.3.4+3.4.5+...+99.100.101-(1.2+2.3+3.4+...+99.100)
Đặt
A=1.2.3+2.3.4+3.4.5+...+99.100.101
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+99.100.101.4=
=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+99.100.101.(102-98)=
=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-98.99.100.101+99.100.101.102=
=99.100.101.102
=> A=99.100.101.102:4=99.25.100.102
Đặt
B=1.2+2.3+3.4+...+99.100
3B=1.2.3+2.3.3+3.4.3+...+99.100.3=
=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)=
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=
=99.100.101
=> B=99.100.101:3=33.100.101
=> S=A-B
Bạn tự tính nốt nhé
b/
Tổng trên có 51 số hạng
A=1+(2+22)+(23+24)+...+(249+250)=
=1+2(1+2)+23(1+2)+...+249(1+2)=
=1+3(2+23+25+...+249) => A:3 dư 1
Ta có
A=(1+2+22)+(23+24+25)+(26+27+28)+...+(248+249+250)=
=7+23(1+2+22)+26(1+2+22)+...+248(1+2+22)=
=7(1+23+26+...+248) chia hết cho 7
Ta có
A=1+2+22+(23+24+25+26)+...+(247+248+249+250)=
=7+23(1+2+22+23)+...+247(1+2+22+23)=
=7+15(23+...+247)
=> A chia 15 dư 7
