CMR: 1 - 1/2+ 1/3 -1/4+......+1/99-100 = 1/101 + 1/102+......+1/200
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QT
0
VD
2
AK
28 tháng 7 2018
P/s : Đề sai mik sửa lại rồi : Tham khảo nhé :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
N
27 tháng 8 2018
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-2.\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-1+\frac{1}{2}+....+\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
NT
0
NM
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PT
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TH
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NM
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LN
22 tháng 3 2016
1.Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
22 tháng 3 2016
1 .Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
S
27 tháng 5 2017
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
HN
27 tháng 5 2017
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
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