Tính a) 50 x 51 + 51 x 52 + 52 x 53 + … + 100 x 101 b) 1 mũ 2 + 2 mũ 2 + 3 mũ 2 + … + 100 mũ 2 11/10/2021 Bởi Audrey Tính a) 50 x 51 + 51 x 52 + 52 x 53 + … + 100 x 101 b) 1 mũ 2 + 2 mũ 2 + 3 mũ 2 + … + 100 mũ 2
`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` $\\$ `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)` $\\$ 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` $\\$ `=>B=C+D=333300+5050=338350` Vậy `1^2+2^2+3^2+…+100^2=338350` Bình luận
`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`
$\\$
`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)`
$\\$
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`
$\\$
`=>B=C+D=333300+5050=338350`
Vậy `1^2+2^2+3^2+…+100^2=338350`