A=1.2+2.3+3.4+....+29.30

3A=1.2.3+2.3.3+3.4.3+....+29.30.3

3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+...+29.30(31-28)

3A=0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+28.29.30-29.30.31

3A=29.30.31=26970

A=8990

A=8990

ko mình mình ko lại

A = 1.2 + 2.3 + 3.4 + ........+ 29.30

3A =1.2.3 + 2.3.3 + .... +29.30.3 

 3A   =1.2.3 + 2.3.(4-1)+ ......+ 29.30.(31-28)

3A = 1.2.3 + 2.3.4 - 1.2.3 +....-28.29.30

A = 28.29.30/3= 8120

a/ \(3A=1.2.3+2.3.3+3.4.3+4.5.3+...+29.30.3.\)

\(3A=1.2.3+2.3\left(4-1\right)+3.4.\left(5-2\right)+4.5\left(6-3\right)+...+29.30\left(31-28\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+29.30.31-28.29.30\)

\(3A=29.30.31\Rightarrow A=\frac{29.30.31}{3}=10.29.31=8990\)

c/ \(C=1+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+...+30\left(29+1\right)\)

\(C=1+2+1.2+2.3+3+3.4+4+...+29.30+30\)

\(C=\left(1+2+3+4+...+30\right)+\left(1.2+2.3+3.4+...+29.30\right)\)

Dấu ngoặc thứ nhất là tính tổng 1 cấp số cộng, dấu ngoặc thứ 2 chính là câu a

b/ Câu b dãy viết ngắn quá chưa tìm ra quy luật

a) A = 1.2 + 2.3 + ... + 29.30

=> 3A = 1.2.3 + 2.3.(4-1) + ... + 29.30.(31-28)

          = 1.2.3 + 2.3.4 - 1.2.3 + ... + 29.30.31 - 28.29.30

          = 29.30.31

=> A = \(\frac{29.30.31}{3}=8990\)

a) \(A=1.2+2.3+3.4+...+29.30\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+29.30\left(31-28\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+29.30.31-28.29.30\)

\(\Rightarrow3A=29.30.31\)

\(\Rightarrow A=29.30.31:3\)

\(\Rightarrow A=29.10.31\)

\(\Rightarrow A=8990\)

3A= 1.2.3+2.3.4+3.4.3 +......+ 29.30.3

3A= 1.2. ﴾3 ‐ 0﴿ + 2.3.﴾4 ‐ 1﴿ +3.4. ﴾5 ‐ 2﴿....... . 29.30. ﴾31 ‐ 28﴿

3A = ﴾1.2.3 + 2.3.4 + 3.4.5 +...... +18.20.21﴿ ‐ ﴾0.1.2 + 1.2.3 + 2.3.4 +.......+ 18.19.20﴿

3A = 29.30.31 ‐ 0.1.2

3A =26970‐0

3A= 26970

 A=26970:3

A = 8990.

Vậy A=8990

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 >:)!!!!!

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

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

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

=>\(A=1-\frac{1}{50}\)

=>\(A=\frac{49}{50}\)

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

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

\(\Rightarrow A=1-\frac{1}{50}\)

\(\Rightarrow A=\frac{49}{50}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

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

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

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

\(A=\frac{49}{50}\)