3S=1.2.3+2.3.3+3.4.3+...+49.50.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50
=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51
=0+0+...+0+0+49.50.51
=49.50.51
S=(49.50.51)/3
=41650
Đáp số:41650

lê thị ngọc anh trả lời đúng đấy tin bạn ấy đi

3S=1.2.3+2.3.3+3.4.3+...+49.50.3

=1.2.3+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50

=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51

=0+0+...+0+0+49.50.51

=49.50.51

S=(49.50.51)/3

=41650

Bài 3:Tổng là:

(98,99-1,2):1,1+1) x (98,99+1,2) : 2 = 4503,5405

Đáp số:4503,5405

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: C = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3C = 3.(1.2 + 2.3 + 3.4 + ..... + 98.99)

=> 3C = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 98.99.(100 - 97)

=> 3C = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100

=> 3C = 98.99.100

=> C = $C=\frac{98.99.100}{3}=485100$

Bài này mình vừa giải :D http://olm.vn/hoi-dap/question/185493.html  -- số khác

Ta có 3 x S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3

3 x S = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + 99 x 100 x (101 - 98)

3 x S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 +  3 x 4 x 5 - 2 x 3 x 4 + .. + 99 x 100 x 101 - 98 x 99 x 100

=> 3 x S = 99 x 100 x 101 

=> A = 33 x 100 x 101 = 333300

=333300 nhe

S = 333300

k mình nhé

S = 1.2 + 2.3 + 3.4 + 4.5 + ..... + 99.100

3S=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+99.100.(101-98)

3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

3S=99.100.101

S=99.100.101/3

S=333300

Đặt tổng trên là A

Có : 3A = 1.2.3+2.3.3+....+n.(n+1).3

= 1.2.3+2.3.(4-1)+......+n.(n+1).[(n+2)-(n-1)]

= 1.2.3+2.3.4-1.2.3+.....+n.(n+1).(n+2)-(n-1).n.(n+1)

= n.(n+1).(n+2)

=> A = n.(n+1).(n+2)/3

Tk mk nha

Đặt A=1.2+2.3+...+n(n+1)

3A=1.2.3+2.3.3+...+n(n+1).3

3A=1.2.(3-0)+2.3.(4-1)+...+n(n+1)[(n+2)-(n-1)]

3A=1.2.3-0.1.2+2.3.4-1.2.3+...+n(n+1)(n+2)-(n-1)n(n+1)

3A=[1.2.3+2.3.4+...+n(n+1)(n+2)]-[0.1.2+1.2.3+...+(n-1)n(n+1)]

3A=n(n+1)(n+2)-0.1.2

3A=n(n+1)(n+2)

A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

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: \(1+2-3-4+...+97+98-99-100\)

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

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

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

100 số hạng đầu là

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)

ta có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}\)\(=\frac{100}{101}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{200}-\frac{1}{201}\)

\(=1-\frac{1}{201}\)

\(=\frac{200}{201}\)

 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{200}-\frac{1}{201}\)

\(=1-\frac{1}{201}=\frac{200}{201}\)

Ủng hộ nha,tớ ko ăn cóp đâu.

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{200.201}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{200}-\frac{1}{201}\)

\(=\frac{1}{1}-\frac{1}{201}\)

\(=\frac{200}{201}\)