c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)

\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)

\(\Leftrightarrow A=33\cdot100\cdot101=333300\)

b) Ta có: \(1+2-3-4+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=-4\cdot25=-100\)

S = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰

2S = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹

S = 2S - S

= (2 + 2² + 2³ + ... + 2¹⁰¹) - (1 + 2 + 2² + ... + 2¹⁰⁰)

= 2¹⁰¹ - 1

------------

S = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101

3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)

= 1.2.3 - 1.2.3 + 2

3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102

= 100.101.102

S = 100 . 101 . 102 : 3

= 343400

------------

Q = 1² + 2² + 3² + ... + 100² + 101²

= 101.102.(2.101 + 1) : 6

= 348551

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 >:)!!!!!

Gọi biểu thức này là S , ta có 

S =1.2 + 2.3 + 3.4 + 4.5 + ...+ 100.101

3S= 1. 2 .3 + 2. 3 .3 + 3 . 4 .3 + 4 .5 .3 + ...........+ 100 .101 .3

3S= 1.2 (3 - 0) + 2 . 3 .(4 - 1) + 3 . 4. (5 - 2 ) +.......+ 100 . 101 . (102 - 99)

3S = 1 . 2 . 3 - 0 . 1 .2  + 2 . 3 . 4 - 1 . 2 .3 + ................+ 100 . 101 .102 - 99  100 . 101 

S   =   \(\frac{100.101.102}{3}=\frac{100.101.34}{1}\)

S   = 343400

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)

\(=1.\frac{1}{101}=\frac{1}{101}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)

\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)

\(=\frac{1}{101}\)

A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100 + 100.101

3.A = 1.2.3 + 2.3.3 +3.4.3 + ... + 100.101.3

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

3A = 1.2.3 + 2.3.4 - 1.2.3 + 2.3.4 -3.4.5 + ... +99.100.101 -100.101.102

3A = 99.100.101

A = 99.100.101 : 3

A = 33.100.101

Vậy A = 33. 100 .101 (Tự tính)

A=1.2+2.3+3.4+.....+99.100+100.101

333300

các bạn giúp mk với

A = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101

⇒ 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)

= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102

= 100.101.102

= 1030200

⇒ A = 1030200 : 3

= 343400

Ko

A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102

A=1-1/102=102/102-1/102=101/102

ý b thì chờ mình tí tìm cách lập luận đã nhé

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)

\(A=1-\frac{1}{102}\)

\(A=\frac{101}{102}\)

B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102

B=1-1/102