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22 tháng 3 2015

A=6+16+30+48+...+19600+19998

2A = 1.3+2.4+3.5+...+99.101 

B=2+5+9+14+...+4949+5049

2A = 1.4+2.5+3.6+...+99.102

C=1.2.3+2.3.4+3.4.5+...+98.99.100

4A = 1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+98.99.100.(101-97)
4A = 1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100
4A = 98.99.100.101

21 tháng 9 2016

A=6+16+30+48+...+19600+19998

A : 2 = 3 + 8 + 15 + 24  + . . . + 9800 + 9999

A : 2 = 1.3 + 2.4 + 3.5 + 4.6 + . . . + 98.100 + 99.101

A : 2 = 1.[1+2] + 2.[1+3] + 3.[1+4] + 4.[1+5] + . . . + 98.[1+99] + 99.[1+100]

A : 2 = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 + 4.5 + . . . + 98 + 98.99 + 99 + 99.100

A : 2 = 1 + 2 + 3 + 4 + . . . + 199 + 1.2 + 2.3 + 3.4 + 4.5 + . . . + 98.99 + 99.100

A : 2 = 4950 + 333300

A = 676500

4 tháng 1 2019

Bạn cho sai đề rồi ! 

Sửa : Chứng tỏ : \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{4949}{9900}\)

Ta có :  \(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

 \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(=\frac{99.100-2}{2.99.100}\)

\(=\frac{4949}{9900}=VP\)

Study well ! >_<

19 tháng 4 2016

= 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + 1/4.5 - ........+1/98.99 - 1/99.100 )

=1/2.(1/1.2 - 1/99.100)

=1/2 . 4949/9900

=4949/19800

13 tháng 1 2023

A= 1.2.3 + 2.3.4 + 3.4.5 +.....+ 98.99.100

4A = 98.99.100.4 + .....+ 3.4.5.4 + 2.3.4.4 + 1.2.3.4

4A = 98.99.100.(101-97) +... + 2.3.4.(5-1) + 1.2.3.4

4A = 98.99.100.101 - 97.98.99.100+......+2.3.4.5 - 1.2.3.4 + 1.2.3.4

4A = 98.99.100.101

  A = 98.99.100.101 : 4

  A = 24497550

2 tháng 4 2017

thứ mấy bn nộp

7 tháng 9 2017

Ta có:

\(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)

\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4\)

\(\Rightarrow4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+98.99.100.\left(101-97\right)\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...98.99.100.101-97.98.99.100\)

\(\Rightarrow4A=98.99.100.101\)

\(\Rightarrow A=\dfrac{98.99.100.101}{4}\)

Vậy \(A=\dfrac{98.99.100.101}{4}\)

7 tháng 9 2017

Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)

\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)

\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)

\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)

\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)

\(4A=98.99.100.101\)

\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101\

4A=98.99.100.101

A=\(\dfrac{\text{98.99.100.101}}{4}\)

tick nha

7 tháng 9 2017

Ta có: \(A=1.2.3+2.3.4+3.4.5+...+98.99.100\)

\(4A=\left(1.2.3+2.3.4+...+98.99.100\right)4\)

\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)...+98.99.100.\left(101-97\right)\)

\(4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)

\(4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+97.98.99.100-97.98.99.100+98.99.100.101\)

\(4A=98.99.100.101\)

\(\Rightarrow A=\dfrac{98.99.100.101}{4}=24497550\)

14 tháng 1 2016

[1.2.3+98.99.100]x2+1=bạn tự tính nha
 

14 tháng 1 2016

đây là bài lớp 5
 

A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)

=1/2*(1/2-1/99*100)

=1/2*(4950-1/9900)

=4950/19800

14 tháng 4 2019

\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)