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Cần phải CM: \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{198.200}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{198.200}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{198}-\frac{1}{200}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{200}\)
\(\Rightarrow A=\frac{99}{200}\)
\(\Rightarrow\frac{1}{2}A=\frac{99}{200}\)
\(\Rightarrow A=\frac{99}{400}\)
Có: \(\frac{1}{4}=\frac{100}{400}\)
Lại có: \(\frac{99}{400}< \frac{100}{400}\)
Vậy A < 1/4 (đpcm)
Dự vào thừa số thứ nhất ở mẫu , ta xác định được thừa số thứ nhất ở mẫu của số hạng thứ 100 là :
\(2+2\left(100-1\right)=200\)
Tức là chứng minh :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{200.202}< \frac{1}{4}\)
Ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{200.202}\)
\(=\frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)< \frac{1}{4}.1=\frac{1}{4}\)
Vậy
Dự vào thừa số thứ nhất ở mẫu, ta xác định thừa số thứ nhất ở mẫu của số hạng thứ 100 là :
\(2+2\left(100-1\right)=200\)
Tức là chứng minh :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{200.202}< \frac{1}{4}\)
Ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{200.202}\)
\(=\frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)< \frac{1}{4}.1=\frac{1}{4}\)
Vậy ...
S = 1/2 . ( 1/2 -1/2 + 1/6 -1/2 + ...+ 1/99 - 1/100)
S= 1/2 . (1-2 - 1/100)
S=1/2 . 49/100
S= 49/200
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
=>2S=\(2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\right)\)
=\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
=>S=\(\frac{49}{100}:2=\frac{49}{100}.\frac{1}{2}=\frac{49}{200}\)
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}< \frac{3}{4}\)
\(\Leftrightarrow A=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\frac{2014}{2015}+\frac{1007}{2016}\)
\(\Leftrightarrow A=1,5\)
Đổi \(\frac{3}{4}=0,75\)
Vì 0,75 < 1,5
Nên ko thể CM
Bài này mà cũng hỏi thì đừng có thi nữa. đợi vài ngày sau có đáp án nhé.
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n.\left(2n+2\right)}\))
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{4}-\frac{1}{2.\left(2n+2\right)}\)
\(=\frac{1}{4}-\frac{1}{4n+4}=\frac{1}{4}-\frac{1}{4.\left(n+1\right)}\)
\(=\frac{n+1}{4.\left(n+1\right)}-\frac{1}{4.\left(n+1\right)}=\frac{n+1-1}{4.\left(n+1\right)}=\frac{n}{4.\left(n+1\right)}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
Có gì ko hiểu thì cứ hỏi mình nha :)
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.2\frac{2}{4}+2.2\frac{2}{4.6}+2.2\frac{2}{6.8}+...+2.2\frac{2}{2008.2010}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}\)
\(=\frac{1004}{1005}\)