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

= 1 - 1/100

= 99/100

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2006.2007}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(=1-\frac{1}{2007}\)

\(=\frac{2006}{2007}\)

CẦN GẤP NHÉ MỌI NGƯỜI

Ta có:

\(A=\frac{1}{1\text{x}2\text{x}3}+\frac{1}{2\text{x}3\text{x}4}+\frac{1}{3\text{x}4\text{x}5}+...+\frac{1}{18\text{x}19\text{x}20}< \frac{1}{4}\)

\(A=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{20}< \frac{1}{4}\)

\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\frac{1}{20}< \frac{1}{4}\)

\(A=1+\frac{1}{20}< \frac{1}{4}\)

\(A=\frac{19}{20}< \frac{1}{4}\)

\(A=\frac{19}{20}< \frac{5}{20}\)

\(A>\frac{1}{4}\)

Cyak 3ampo

+Câu a:

A = 1/1.2 + 1/2.3 + ...+ 1/5.6 + 1

A = 1/1 - 1/2 + 1/2 - 1/3 + ...+ 1/5 - 1/6 + 1

A = 1/1 - 1/6 + 1

A = 6/6 - 1/6 + 6/6

A = 5/6 + 6/6

A = 11/6

Câu b:

B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100

B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)

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

B = 1/2.(1/2 - 1/9900)

B = 1/2.4949/9900

B = 4949/19800

Câu a:

1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1

= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1

= 1/1 - 1/6 + 1

= 6/6 - 1/6 + 6/6

= 5/6 + 1

= 11/6

Câu b:

Câu b:

B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100

B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)

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

B = 1/2.(1/2 - 1/9900)

B = 1/2.4949/9900

B = 4949/19800

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)(ĐPCM)

\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=190/380-1/380

=189/380

Gọi biểu thức trên là S. Ta có :

\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)

Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)

Ta sẽ có : 

\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)

\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)

`=1/2(1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/18×19 - 1/19×20)`
`=1/2(1/2 - 1/19×20)`
`=1/2×189/380 `
`=189/760`

Câu a:

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

3A = 1.2.3 + 2.3.3 +..+ 99.100.3

1.2.3 = 1.2.3

2.3.3 = 2.3(4 - 1) = 2.3.4 - 1.2.3

3.4.4 = 3.4(5 - 2) = 3.4.5 - 2.3.4

.............................................................

99.100.3 = 99.100.(101 - 98)=99.100.101-98.99.100

Cộng vế với vế ta có:

3A = 99.100.101

A = 99.100.101 : 3

A = 333300

Câu b:

1.3 + 3.5 + 5.7 + ..+ 97.99

6.A = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 97.99.6

1.3.6 = 1.3.(5 + 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5.(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7

.......................................................

97.99.6 = 97.99.(101 - 95) = 97.99.101-95.97.99

Cộng vế với vế ta có:

6A = 1.3.1 + 97.99.101

A = (1.3.1 + 97.99.101) : 6

A = (3 + 9603.101) : 6

A = (3+ 969903) : 6

A = 969906 : 6

A = 161651