Tính tổng dãy số sau : \(A=\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+....+\frac{5}{9506}+\frac{5}{9702}+\frac{5}{9900}\)
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PM
4
28 tháng 5 2018
A = 5 x (\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\))
A = 5 x ( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\))
A = 5x( \(\frac{1}{2}-\frac{1}{100}\))
A = \(\frac{49}{20}\)
28 tháng 5 2018
Gọi tổng trên là A
\(\Leftrightarrow A=5\times\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)
(Tính dãy trong ngoặc) Gọi dãy trong ngoặc là B
\(\Leftrightarrow2B=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)
\(\Leftrightarrow2B-B=\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)
\(\Leftrightarrow B=\frac{1}{3}-\frac{1}{9900}+0+...+0\)
\(\Leftrightarrow B=\frac{3299}{9900}\)
\(\Rightarrow A=5\times\frac{3299}{9900}\)
\(\Rightarrow A=1,6661616...\approx1,7\)
T
2 tháng 9 2020
\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)
\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)
TK
3
3 tháng 5 2017
Hơi nhầm nè , để tôi sửa lại đề \(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\)
\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(A=1+1+1+...+1-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-....-\frac{1}{9900}\)
\(A=98-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\right)\)
\(A=98-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(A=98-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=98-\left(\frac{1}{2}-\frac{1}{100}\right)=98-\frac{49}{100}=\frac{9751}{100}\)
Vậy.............
S
3 tháng 5 2017
\(A=\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9989}{9900}\)
\(A=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
có 50 số 1
\(A=50-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
Đặt B = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
Thay B vào A ta được:
\(A=50-\frac{49}{100}=\frac{5000}{100}-\frac{49}{100}=\frac{4951}{100}\)
T
0
25 tháng 4 2018
Ta có :
\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)
\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)
\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)
\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)
\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)
Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra :
\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{A}{100}=98-\frac{49}{100}\)
\(\frac{A}{100}=\frac{9751}{100}\)
\(A=\frac{9751}{100}.100\)
\(A=9751\)
Vậy \(A=9751\)
Chúc bạn học tốt ~
25 tháng 5 2016
A=1+2+3+4+5+...+99+100
A=(1+100).100:2=101.50=5050
B=1/2+1/6+1/12+1/20+1/30+...+1/9900
B=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100
B=1-1/100=99/100
25 tháng 5 2016
A = 100 x 101 : 2 = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
1 tháng 8 2020
\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)
1 tháng 8 2020
Trả lời:
2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}\)
\(K=\frac{49}{100}\)
3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\frac{102}{103}\)
\(L=\frac{170}{103}\)
Học tốt
A= 5.(1/2 + 1/6+1/12+1/20+...+1/9506+1/9702+1/9900)
= 5. (1/1.2 + 1/2.3+1/3.4+1/4.5+...1/97.98+1/98.99+1/99.100)
= 5 .(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/97-1/98+1/98-1/99+1/99-1/100)
= 5.(1-1/100)=5. 99/100=99/20
25
5144 nhe