Gọi A là biểu thức ta có: 
A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

1.2 + 2.3 + 3.4 +...+99.100 = \(\frac{99.\left(99+1\right)\left(99+2\right)}{3}=333300\)

cặp hai số đầu và cuối, cg như từng cặp số cách đều số đầu và cuối đều có tổng =101

có 50 cặp như thế, do đó kết quả là 101.50=5050

\(G=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)

\(=1\left(1+1\right)+2\left(1+2\right)+3\left(1+3\right)+...+99\left(1+99\right)\)

\(=\left(1^2+2^2+3^3+...+99^2\right)+\left(1+2+3+...+99\right)\)

\(=\dfrac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\dfrac{99\cdot100}{2}\)

=328350+4950

=333300

1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/x - 1/x+1 = 99/100

1- 1/x+1= 99/100

1/x+1= 1- 99/100

1/x+1=1/100

=> x+1 = 100

     x= 100-1

     x=99

4/1.2 + 4/2.3 +4/3.4+...+4/99.100

=> 4/1 - 4/2 + 4/2 - 4/3 + 4/3 - 4/4+...+ 4/99 - 4/100

=>4 - 4/100

\(\frac{4}{1.2}+\frac{2}{2.3}+\frac{4}{3.4}+...+\frac{4}{99.100}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=4.\left(1-\frac{1}{100}\right)\)

\(=4.\frac{99}{100}=\frac{99}{25}\)

3S = 1.2.3 + 2.3.3 + 1.4.3 + ... + 99.100.3

     = 1.2.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

     = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 3.4.2 + .. + 99.100.101 - 99.100.98

   => 99.100.101  => S = 99.100.101/3 = 999900

 Vậy:  S = 999900

đề gì mà nhân rồi cộng

ai tính đuoccưẹ

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: \(1+2-3-4+...+97+98-99-100\)

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

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

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

a) \(1.2+2.3+3.4+...+19.20\)

\(=\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)

\(=3080\)

b) \(9+99+999+...+999...9\left(100so9\right)\)

\(\)\(=\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+...+\left(1000...0-1\right)\left(99so0\right)\)

\(=\left(10+10^2+10^3+...10^{99}\right)+\left(-1\right).100\)

\(=\left(1+10+10^2+10^3+...10^{99}\right)+\left(-1\right).101\)

\(=\dfrac{10^{99+1}-1}{99-1}-101\)

\(=\dfrac{10^{100}-1}{98}-101\)

\(=\dfrac{10^{100}-9899}{98}\)

c) \(999.9x222...2\) (100 số 9; 100 số 2)

\(9x2=18\)

\(99x22=2178\)

\(999x222=\text{221778}\)

\(9999x2222=22217778\)

\(99999x22222=2222177778\)

\(.........\)

Theo quy luật trên ta có 100 số 9 nhân 100 số 2:

\(999.9x222...2=222...21777...78\) (99 sô 2; 1 số 1; 99 số 7; 1 số 8)

A, 1.2 + 2. 3 + 3. 4 + ....+ 19 . 20 

⇒\(\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)

⇒3080 

vậy kết quả câu a, là 3080 

A=9/1.2+ 9/2.3+ 9/3.4+ .... +9/98.99 + 9/99/100

  =9(1- 1/2 + 1/2 -1/3+...+1/99 -1/100)

  =9.(1- 1/100)

  =9.99/100

  =891/100

A=9/1.2+9/2.3+...+9/99.100

A/9=1/1.2+1/2.3+....+1/99.100

A/9=1-1/2+1/2-1/3+....+1/99-1/100

A/9=1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/99+1/99)-1/100

A/9=1-1/100

A/9=99/100

A=99/100.9=891/100

     Vậy A=891/100

 mik ko biết đúng hay sai mn góp ý giúp mik nha

\(A=\frac{9}{1\cdot2}+\frac{9}{2\cdot3}+\frac{9}{3\cdot4}+...+\frac{9}{98\cdot99}+\frac{9}{99\cdot100}\)

\(\Rightarrow A=9\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)

\(\Rightarrow A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=9\left(1-\frac{1}{100}\right)=9\cdot\frac{99}{100}=\frac{891}{100}\)

HK TỐT #