1/2 + 1/6 + 1/12 +...+ 1/9900 + 1/10100
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100
= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101
= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101
= 1/1 - 1/101
= 100 /101
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
=\(1-\frac{1}{101}\)
=\(\frac{100}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1+ 1+1+ ...+ 1 +(\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}+\dfrac{1}{10100}\))
=(1+1+1+...+1)+ (\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}+\dfrac{1}{100x101}\))
=100 +\(1-\dfrac{1}{101}=100-\dfrac{100}{101}=\dfrac{10000}{101}\)
1+1/2+1+1/6+1+1/12+...+1+1/9900
=1+1/1*2+1+1/2.3+....+1+1/99*100
=100*1+1-1/2+1/2-1/3+1/3-1/4...+1/99-1/100
=100+99/100
=10099/100
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{10100}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{100}-\frac{1}{101}\)
=\(1-\frac{1}{101}\)
=\(\frac{100}{101}\)
b,\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)
=\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)\)
=\(1-\frac{1}{512}\)
=\(\frac{511}{512}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) trieu dang làm rồi
b) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2+1/6+1/12+.......+1/10100
1/1x2+1/2x3+1/3x4+...........+1/100x101
1-1/2+1/2-1/3+1/3-..........+1/100-1/101
1-1/101
=100/101
cho bạn công thức mẫu trừ đi bao nhiêu thì tử là bấy nhiêu
vd 2/2x4=1/2-1/4
chúc bạn học tốt
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..+\frac{1}{9900}+\frac{1}{10100}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}+\frac{1}{100\times101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/2+1/6+1/12+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100
=99/100
![](https://rs.olm.vn/images/avt/0.png?1311)
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +....+1 /99.100
= 1/1 - 1/2 + 1/2 -1/3 + .... + 1/99 - 1/100
= 1/1 - 1/100
= 100/100 - 1/100
= 99/100
1/2+1/6+1/12+1/20+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
1/1x2+1/2x3+1/3x4+...+1/99x100+1/100x101
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100+1/100-1/101
=1/1-1/101
=100/101