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Chứng tỏ rằng:
\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{149}+\frac{1}{150}>\frac{5}{6}\)
a) \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< 1\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)+...+\left(\frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}\right)\)\(\frac{1}{60}\cdot10< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}\cdot10\)
\(\frac{1}{6}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{5}\)(1)
\(\frac{1}{70}\cdot10< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{60}\cdot10\)
\(\frac{1}{7}< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{6}\)(2)
.... (tương tự )
\(\frac{1}{100}\cdot10< \frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}< \frac{1}{90}\cdot10\)
\(\frac{1}{10}< \frac{1}{91}+...+\frac{1}{100}< \frac{1}{9}\)
Từ (1)(2)(3)(4) và (5)
\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)
\(\frac{1}{2}< \frac{1624}{2520}< \frac{1}{51}+...+\frac{1}{100}\)
\(1>\frac{1879}{2520}>\frac{1}{51}+...+\frac{1}{100}\)
ta có 1/51>1/100
1/52>1/100
..................
1/100=1/100
\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2
\(\Rightarrow\)S>\(\frac{1}{2}\)
cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá
chúc bạn học tốt~
a=\(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{100}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
=>b/a=2011
hình như đề : CMR : \(\frac{b}{a}\)là 1 số nguyên
Ta có :
\(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(a=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(a=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(a=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(b=\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}\)
\(b=2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{b}{a}=\frac{2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=2011\)là 1 số nguyên ( đpcm )
1/1 . 2 + 1/ 3 . 4 + 1/5 . 6 + ...+ 1/99 . 100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/99 - 1/100
= ( 1 + 1/3 + 1/5 + ...+ 1/99 ) - ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - 2 . ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - ( 1 + 1/2 + ...+ 1/50 )
= 1/51 + 1/52 + ...+ 1/100
Tham khảo nha !!!
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) (đpcm)
ai giúp mk ik
mk đg cần gấp,còn nhìu đề chx lm
ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\) \(-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
Ta có:
1/51 > 1/75
1/52 > 1/75
…
1/75 = 1/75
=> 1/51 + 1/52 + … + 1/75 > 1/75 + 1/75 + … + 1/75 (25 số 1/75)
=> 1/51 + 1/52 + … + 1/75 > 25/75
=> 1/51 + 1/52 + … + 1/75 > 1/3
Ta có:
1/71 > 1/100
1/72 > 1/100
…
1/00 = 1/100
=> 1/71 + 1/72 + 1/73 + … + 1/100 > 1/100 + 1/100 + … + 1/100 ( 25 số 1/100)
=> 1/71 + 1/72 + 1/73 + … + 1/100 > 25/100
=> 1/71 + 1/72 + 1/73 + … + 1/100 > 1/4
Ta có:
1/71 + 1/72 + 1/73 + … + 1/100 > 1/4 (cmt)
1/51 + 1/52 + … + 1/75 > 1/3 ( cmt )
=> 1/51 + 1/52 + … + 1/75 + 1/71 + 1/72 + 1/73 + … + 1/100 > 1/3 + 1/4
=> A > 7/12 (đpcm)