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

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

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

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

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

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

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

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

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

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

/hoi-dap/question/115788.html Giúp mik vs\

mk k vt lại đề nha

S=2.(1/1.2+1/2.3+1/3.4+............+1/99.100)

S=2.(1-1/2+1/3-1/4+1/4-1/5+.............+1/99-1/100)

S=2.(1-1/100)

S=2.99/100

S=198/100

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

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

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

S=\(\frac{2}{1}\).(\(\frac{1}{1}\)-\(\frac{1}{100}\))

S=\(\frac{2}{1}\).(\(\frac{100}{100}\)-\(\frac{1}{100}\))

S=\(\frac{2}{1}\).\(\frac{99}{100}\)

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

Vậy S=\(\frac{99}{50}\)

Đặt tổng trên là A , ta có :

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

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

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

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

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

Bạn có thể làm như vầy nè:

Đặt 2 ra ngoài,ta có dạng S = 2 x (1/2.3 + 1/3.4 + ... + 1 x 98 x 99 + 1/99.100)

Với chú ý:1/2.3 = 1/2 - 1/3

1/3.4 = 1/3 - 1/4,........

Vậy S = 2 x ( 1/2 - 1/100)  = 2 x (50/100 - 1/100) = 2.49/100 = 98/100 = 49/50

Chúc bạn học thiệt là giỏi!

\(Tac\text{ó}:\frac{2}{1.2}-\frac{2}{2.3}-\frac{2}{3.4}-...-\frac{2}{98.99}-\frac{2}{99.100}\)

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

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

Sorry mình ấn nhầmmmmm

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

BÀI 1:

\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(S=1+\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.4}+\frac{1}{4.8}\)

\(S=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\)

\(S=1+1-\frac{1}{8}\)

\(S=\frac{15}{8}\)

BÀI 2:

\(A=1.2+2.3+3.4+...+98.99\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+98.99.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99\)

\(3A=\left(1.2.3+2.3.4+3.4.5+98.99.100\right)-\left(1.2.3+2.3.4+...+97.98.99\right)\)

\(3A=98.99.100\)

\(3A=970200\)

\(\Rightarrow A=970200:3\)

\(A=323400\)

CHÚC BN HỌC TỐT!!!
 

3A=1.2.3+2.3.(4-1)+.............+98.99.(100-97)+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+...........+98.99.100-97.98.99+99.100.101-98.99.100

3A=99.100.101

A=99.100.101:3

A=333300

Ta có : 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 98.99.3 + 99.100.3

=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + 98.99.( 100 - 97 ) + 99.100.( 101 - 98 )

=> 3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100

=> 3A = ( 1.2.3 + 2.3.4 + 3.4.5 + ..... + 98.99.100 + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ..... + 98.99.100 )

=> 3A = 99.100.101 - 0.1.2

=> 3A = 99.100.101

=> A = 33.100.101

=> A = 333300

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

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

3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3

3A= 1.2.3+2.3﴾4‐1﴿+3.4﴾5‐2﴿+...+98.99﴾100‐97﴿+99.100﴾101‐98﴿  

3A= 1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+...‐97.98.99+99.100.101‐98.99.100

3A = 99.100.101 3S = 3.33.100.101

A=33.100.101= 333300

1.2 là 1 x 1 hay 1,2 vậy