Chứng tỏ \(\frac{A}{B}\) là một số nguyên biết :
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....+\frac{1}{2013\cdot2014}\)
\(B=\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\frac{1}{2010\cdot2012}+....+\frac{1}{2014\cdot1008}\)
\(B=\frac{1-3}{1\cdot3}+\frac{2-4}{2\cdot4}+\frac{3-5}{3\cdot5}+\frac{4-6}{4\cdot6}+............+\frac{2011-2013}{2011.2013}+\frac{2012-2014}{2012\cdot2014}-\frac{2013+2014}{2013\cdot2014}\)
Chứng tỏ rằng \(\dfrac{A}{B}\) là một số nguyên biết :
\(B=\dfrac{1}{1008\cdot2014}+\dfrac{1}{1009\cdot2013}+\dfrac{1}{1010\cdot2012}+...+\dfrac{1}{2014\cdot1008}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{2013\cdot2014}\)
\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{2011\cdot2012\cdot2013\cdot2014}\)
1/1.2.3.4+1/2.3.4.5+....+1/2011.2012.2013.2014
Gọi dãy số trên là A , ta có :
3A=3/1.2.3.4+3/2.3.4.5+....+3 / 2011.2012.2013.2014
3A=1/1.2.3-1/2.3.4+1/2.3.4-......-1/2011.2012.2013+1 / 2011.2012.2013 - 1/2012.2013.2014
3A=1/1.2.3-1/2012.2013.2014
phần còn lại bn tính tiếp đi
so sánh A=\(\frac{1}{1\cdot2\cdot3}\)+\(\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2012\cdot2013\cdot2014}\) với \(\frac{1}{4}\)
Lời giải:
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2014-2012}{2012.2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{2012.2013}-\frac{1}{2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2013.2014}< \frac{1}{2}$
$\Rightarrow A< \frac{1}{2}:2$
Hay $A< \frac{1}{4}$
cho hỏi : A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
B=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
tính B-A
giúp vs nha,cảm ơn nhìu!
A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)
B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
B-A=1/1008
cho hỏi: A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2015\cdot2016}\)
B=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Tính B-A
Ai giúp cảm ơn!
Tính
A=\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{2012\cdot2014}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2012.2014}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2014}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\cdot\frac{503}{1007}\)
\(\Leftrightarrow A=\frac{503}{2014}\)
= 1/2[1/2 - 1/4+1/4-1/6 + 1/6-1/8+...+ 1/2012-1/2014]
= 1/2[1/2-1/2014]
= 1/2 * 503/1007
= 503/2014
\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{2012\cdot2014}\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2014}\right)\)
\(\Rightarrow A=\frac{1}{2}\cdot\frac{503}{1007}\)
\(\Rightarrow A=\frac{503}{2014}\)
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
Chứng tỏ rằng:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)