Đặt A=1²+2²+3²+.....+100²

=>A=1.1+2.2+3.3+....+100.100

=>A=1.(0+1)+2.(1+1)+3.(1+2)+....+100.(1+99)

=>A=1.0+1.1+2.1+1.2+3.1+2.3+......+100.1+99.100

=>A=1+2+1.2+3+2.3+......+100+99.100

=>A=(1+2+3+....+100)+(1.2+2.3+....+99.100)

Đặt S=1.2+2.3+....+99.100

=>3S=1.2.3+2.3.3+....+99.100.3

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

=>3S=1.2.3-0.1.2+2.3.4-1.2.3+......+99.100.101-98.99.100

=>3S=1.2.3+2.3.4-1.2.3+.....+99.100.101-98.99.100

=>3S=99.100.101=>\(S=\frac{99.100.101}{3}=333300\)

Đặt P=1+2+3+....+100

từ 1->100 có:100-1+1=100(số hạng)

=>\(1+2+3+....+100=\frac{100.\left(100+1\right)}{2}=5050\)

Vậy A=S+P=333300+5050=338350