B=(1+1/2)+(1+1/6)+91+1/12)+(1+1/9900)
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\(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)\(=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...\left(\frac{1}{99}-\frac{1}{99}\right)=\left(\frac{100}{100}-\frac{1}{100}\right)+0+...+0=\frac{99}{100}\)Vậy B=99/100
MK k chắc nữa
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{100-1}{100}=\frac{99}{100}\)
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B=\(\frac{1}{2.x}+\left(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}...\frac{1}{99.100}\right)\)
=\(\frac{1}{2.x}+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=2\)
=\(\frac{1}{2.x}+\left(1-\frac{1}{100}\right)\)\(=2\)
=\(\frac{1}{2.x}+\frac{99}{100}\)\(=2\)
=\(\frac{1}{2.x}=2-\frac{99}{100}\)
=\(\frac{1}{2.x}=\frac{101}{200}\)
=\(2.x=200\)
=\(x=200:2=100\)
1/2 * x + 1/2 + 1/6 + 1/12 + .... + 1/9900 = 2
<=> 1/2 * x + ( 1/2 + 1/6 + 1/12 + ... + 1/9900 ) = 2
<=> 1/2 * x + ( 1 /1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 ) = 2
<=> 1/2 * x + ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 ) = 2
<=> 1/2 * x + ( 1 - 1/100 ) = 2
<=> 1/2 * x + ( 100/100 - 1/100 ) = 2
<=> 1/2 * x + 99/100 = 2
<=> 1/2 * x = 2 - 99/100
<=> 1/2 * x = 101/100
<=> x = 101/100 : 1/2
<=> x = 101/100 * 2
<=> x = 101/50
Vậy x = 101/50
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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,\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)
=\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)\)
=\(1-\frac{1}{512}\)
=\(\frac{511}{512}\)
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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)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
B = 1/1.2 + 1/2.3 + ......+ 1/99.100
B = 1 - 1/2 + 1/2 - 1/3 +..........+ 1/99 - 1/100
B = 1 - 1/100
B = 99/100 nhé!
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Giải
\(A=1+2+3+4+5+...+99+100\)
Số số hạng của A là: \(\left(100-1\right)\div1+1=100\)(số hạng)
Tổng A là: \(\frac{\left(100+1\right)\times100}{2}=5050\)
Vây A=5050
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(B=\frac{99}{100}\)