Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

cho bài kham khảo nè :

A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3};B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)

Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B

thank nha

A=1.2+2.3+3.4+...+2017.2018

3A=1.2.3+2.3.3+3.4.3+...+2017.2018.3

3A=1.2.3+2.3.(4−1)+3.4.(5−2)+...+2017.2018.(2019−2016)

3A=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+2017.2018.2019−2016.2017.2018

⇒3A=2017.2018.2019

⇒A=2017.2018.20193

A=2017.2018.20193;B=201833=2018.2018.20183

A=2739315938;B=2739316611

⇒A<B

\(A=1.2+2.3+3.4+4.5+............+2017.2018\)

\(3A = 1.2.3 + 2.3.4 +..............+ 2017.1018.3\)

\(3A = 1.2.3 + 2.3.(4-1) + .............. + 2017.2018.(2019-2016)\)

\(3A = 1.2.3 + 2.3.4 - 1.2.3 + ............. + 2017.2018.2019 - 2016.2017.2018\)

\(3A = 2017.2018.2019\)

\(A = \frac{2017.2018.2019}{3}\)

\(B =\frac {2018^3}{3}\)

đến đây ko bt lm

​A rê. Lớp 6 ngược mà hỏi bài đó hở

đây là bài của bảo trân

Nguyễn Huy Thắng hok lớp 9 ak

Ta có : A=1.2+2.3+3.4+....+2015.2016

=>3A= 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 2017.2018.3

=>3A= 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5-2 ) + 4.5.( 6-3 ) + ... 2017 . 2018 . ( 2019 - 2016 )

=>3A=-1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 +.....+ 2017 . 2018 .2019 - 2017 . 2018 . 2016

=>A= 2017 . 2018 . 2019
 

Lời giải :

Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101

3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3

=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)

=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102

=100.101.102

S=100.101.34=343400

1.Tính 

a) Ta có: 

  A=(1-1/2²).(1-1/3²)...(1-1/100²)

=>A=3/2².8/3².....9999/100²

=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)

=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)

=>A=1/100.101/2

=>A=101/200

b) Ta có: 

  B=-1/1.2-1/2.3-1/3.4-...-1/100.101

=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)

=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)

=>B=-(1-1/101)

=>B=-100/101

 c) Ta có:

 C=1.2+2.3+3.4+...+100.101

       =>3C=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3C=100.101.102

       =>3C=1030200

       =>C=343400

Chúc bạn hok tốt nhé >:)!!!!!

A=1.2+2.3+3.4+4.5+...+2017.2018

=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3

3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018

3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)

=> 3A=2017.2018.2019  => \(A=\frac{2017.2018.2019}{3}\);  \(B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)

Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018

=> 2017.2018.2019<2018.2018.2018

=> A<B

Bui The Hao lam dung roi

mk cung dang can bai nay

Thanks vi da dang honganh

\(A=1.2+2.3+...+18.19\)

\(3A=1.2\left(3-0\right)+2.3\left(4-1\right)+...+18.19\left(20-17\right)\)

\(3A=1.2.3-0.1.2+2.3.4-1.2.3+...+18.19.20-17.18.19\)

\(3A=18.19.20=6840\)

\(\Rightarrow A=\dfrac{6840}{3}=2280\)

=2919