A = 1.2+2.3+3.4+4.5+...+98.99+99.100

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

3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3A = 99.100.101

3A = 999900

A = 333300

nhấn đúng cho mk nha!!!!!!!!!!!!

Cái này trên mạng có            

Áp dụng công thức ta có :

\(A=1.2+2.3+3.4+...+99.100=\frac{99.100.101}{3}=333300\)

A=1.2+2.3+3.4+4.5+.....+98.99+99.100 Rút gọn đi ta còn:

A=1+100

=>A=101

Ta có : A = 1.2 + 2.3 + 3.4 + ....... + 99.100

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101

=> 3A = 99.100.101

=> A = 99.100.101/3

=> A = 333300

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

Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 3S = 3.33.100.101
S=33.100.101= 333300

Đặt tổng trên = A

Có : 3A = 1.2.3+2.3.3+....+98.99.3

 = 1.2.3+2.3.(4-1)+.....+98.99.(100-97)

 = 1.2.3+2.3.4-1.2.3+.....+98.99.100-97.98.99

 = 98.99.100

=> A = 98.99.100/3 = 323400

k mk nha

Gọi A = 1.2 + 2.3 + .. + 98.99

3A = 1.2.3 + 2.3.3 + ... + 98.99.3

3A = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99

3A = 98.99.100

3A = 970200

A = 323400

đặt A = 1.2 + 2.3 + 3.4 + ... + 97.98 + 98.99

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 97.98.3 + 98.99.3

3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + 97.98.(99-96) + 98.99.(100-97)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 97.98.99 - 96.97.98 + 98.99.100 - 97.98.99

3A = 98.99.100

A = 98,99.100 : 3

A = 333300

Đặt tổng trên là A , ta có :

\(\frac{A}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\frac{A}{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{A}{2}=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)\(\frac{A}{2}=\frac{99}{100}\)

\(A=\frac{99}{100}.2\)

\(A=\frac{99}{50}\)

\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)

\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(S=2\times\left(1-\frac{1}{100}\right)\)

\(S=2\times\frac{99}{100}\)

\(S=\frac{99}{50}\)

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)

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