Chứng Minh:
A = \(\frac{1}{3}\)+ \(\frac{1}{15}\) + \(\frac{1}{35}\)+ ... +\(\frac{1}{9999}\) +\(\frac{1}{10403}\) < \(\frac{1}{2}\)
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Những câu hỏi liên quan
NM
1
PY
2
BN
22 tháng 5 2016
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{98}{303}\)
\(A=\frac{49}{303}\)
22 tháng 5 2016
A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A=\(\frac{98}{303}.\frac{1}{2}\)
A=\(\frac{49}{303}\)
Chúc bạn học tốt!
LH
2
NT
10
IW
9 tháng 7 2016
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)
\(\Rightarrow A=\frac{44}{303}\)
MH
9 tháng 7 2016
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
=> A = 98/203 : 2 = 49/303
MP
25 tháng 1 2016
A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
DD
5 tháng 1 2016
\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)
A.2=2/3.5+2/5.7+2/7.9+…+2/99.101
A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101
Vậy
A.2=1/3-1/101=98/303
A=98/303/2=49/303
Đúng không
4 tháng 1 2016
A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101
= 1/3 - 1/101 = 98/303
Vậy A = 98/303 : 2 = 49/303
VN
6
LA
27 tháng 6 2017
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=\(1-\frac{1}{101}=\frac{100}{101}\)
19 tháng 4 2015
1.
= \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.............+\frac{1}{99.101}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\)
= 1 - \(\frac{1}{101}\)
= \(\frac{100}{101}\)
25 tháng 12 2016
siêu tốc
\(2000+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=2000+\frac{50}{3.101}\)
LA
25 tháng 12 2016
Ta có: \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
=\(2000+\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
Đặt A=\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
=> 2xA =\(\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{99x101}\right)\)
2xA = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
2xA = \(\frac{1}{3}-\frac{1}{101}\)
2xA = \(\frac{98}{303}\)
A = \(\frac{98}{606}=\frac{49}{303}\)
=> \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)=2000+\frac{49}{303}=\frac{606049}{303}\)
Bạn xem rút gọn được thì rút nhé
HT
0
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{10403}\)
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{101.103}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{101.103}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}+\frac{1}{103}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{103}\right)\Rightarrow A=\frac{1}{2}.\frac{100}{309}\)
\(\Rightarrow A=\frac{50}{309}< \frac{50}{100}< \frac{1}{2}\)
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}+\frac{1}{10403}.\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}+\frac{2}{101.103}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\)
\(2A=1-\frac{1}{103}\)
\(2A=\frac{102}{103}\)
\(A=\frac{51}{103}\)
Ta có : \(\frac{51}{103}< \frac{1}{2}\)
\(\Rightarrow\)\(A< \frac{1}{2}\)