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7 tháng 3 2018

BÀI 1:

\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)

\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)

\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(N=2.\frac{9}{20}\)

\(N=\frac{9}{10}\)

BÀI 2:

\(C=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)

\(3B=99.100.101\)

\(3B=999900\)

\(\Rightarrow B=999900:3\)

\(B=333300\)

CHÚC BN HỌC TỐT!!!!

8 tháng 3 2018

Thanks CÔNG CHÚA ÔRI vì đã giúp mk

4 tháng 8 2016

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

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

A = \(1-\frac{1}{100}=\frac{99}{100}\)

20 tháng 7 2016

\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)

\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=3.\left(1-\frac{1}{10}\right)\)

\(A=3.\frac{9}{10}=\frac{27}{10}\)

\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)

\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)

20 tháng 7 2016

a) Lấy A chia 3

b) Lấy B nhân 3/2

16 tháng 2 2019

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

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

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

  = \(1-\frac{1}{100}\)

  = \(\frac{99}{100}\)

Vậy ...

16 tháng 2 2019

B = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

   = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

   = \(\frac{1}{2}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{20}\)

  = \(\frac{1}{2}-\frac{1}{20}\)

  = \(\frac{9}{20}\)

Vậy B = 9/20

27 tháng 1 2018

S=1.2+2.3+3.4+....+99.100

3S = 1.2.3 + 2.3.3 + 3.4.3+...... + 99.100.3

3S = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +.....+  99.100.(101-98)

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

3S = 99.100.101

S = \(\frac{99.100.101}{3}=333300\)

27 tháng 1 2018

A = 1.2+2.3+3.4+......+99.100 

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

6 tháng 5 2018

Bài 1

a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{33}{99}-\frac{1}{99}\)

\(=\frac{32}{99}\)

c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Tk mình nha!!

7 tháng 5 2018

Câu 2:

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

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

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)

\(=\frac{3\cdot100}{2}\)

\(=\frac{300}{2}=150\)

25 tháng 7 2016

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\\ =\left(\frac{2}{1}-\frac{2}{2}\right)+\left(\frac{2}{2}-\frac{2}{3}\right)+...+\left(\frac{2}{5}-\frac{2}{6}\right)\\ =\frac{2}{1}+\left(\frac{2}{2}-\frac{2}{2}\right)+....+\left(\frac{2}{5}-\frac{2}{5}\right)-\frac{2}{6}\\ \)

\(=\frac{2}{1}-\frac{2}{6}\\ =\frac{10}{6}=\frac{5}{3}\)

Câu b và c tương tự như câu a 

d) Số hạng thứ 225 của dãy là : 

5+(225-1).5=1125

Tổng của dãy trên là : 

(1125+5).225:2=127125

e) Số số hạng là : 

(2016-2):2+1=1008 ( số hạng ) 

Tổng của dãy trên là : 

(2016+2).1008:2=1017072

25 tháng 7 2016

a)\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)

\(=2.\left(1-\frac{1}{6}\right)\)

\(=2.\frac{5}{6}\)

\(=\frac{5}{3}\)

b)\(\frac{2}{1.3}+\frac{2}{5.3}+\frac{2}{5.7}+\frac{2}{9.7}+...+\frac{2}{2015.2017}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

c)\(\frac{12}{2.5}+\frac{12}{8.5}+\frac{12}{11.8}+\frac{12}{14.11}+\frac{12}{17.14}+\frac{12}{20.17}\)

\(=4.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=4.\frac{9}{20}\)

\(=\frac{9}{5}\)

 

14 tháng 3 2017

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

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

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

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

\(=2.\frac{99}{100}=\frac{99}{50}\)

14 tháng 3 2017

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

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

=\(2\cdot\frac{99}{100}=\frac{99}{50}\)

27 tháng 8 2017

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

27 tháng 8 2017

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)