`a)` Đặt $A=50.51+51.52+52.53+...+100.101$
`=>A.3=50.51.3+51.52.3+52.53.3+...+100.101.102.3`
`=>3A=50.51.(52-49)+51.52.(53-50)+52.53.(54-51)+...+100.101.(102-99)`
`=>3A=50.51.52-49.50.51+51.52.53-50.51.52+52.53.54-51.52.53+...+100.101.102-99.100.101`
`=>3A=-49.50.51+(50.51.52-50.51.52)+(51.52.53-51.52.53)+...+(99.100.101-99.100.101)+100.101.102`
`=>3A=100.101.102-49.50.51`
`=>3A=2.50.101.2.51-49.50.51`
`=>3A=50.51.(4.101-49)=50.3.17.355`
`=>A={50.3.17.355}/3=50.17.355=301750`
Vậy: `50.51+51.52+52.53+...+100.101=301750`
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`b)` Ta có:
`n^2-n=n(n-1)`
`=>n^2=n(n-1)+n` (*)
Áp dụng (*)
Đặt `B=1^2+2^2+3^2+...+100^2`
`=>B=1.0+1+2.1+2+3.2+3+...+100.99+100`
`=>B=(1.2+2.3+...+99.100)+(1+2+3+...+100)`
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Đặt `C=1.2+2.3+...+99.100`
`=>3C=1.2.3+2.3.3+...+99.100.3`
`=>3C=1.2.3+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=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...+(98.99.100-98.99.100)+99.100.101`
`=>3C=99.100.101`
`=>C={99.100.101}/3=33.100.101=333300`
$\\$
Đặt $D=1+2+3+...+100$
Tổng $D$ gồm có: `{100-1}/1+1=100` số hạng
`=>D={(100+1).100}/2=5050`
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`=>B=C+D=333300+5050=338350`
Vậy `1^2+2^2+3^2+...+100^2=338350`
