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\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\right)\)
\(M=\frac{1}{2}.\left(1-\frac{1}{102}\right)\)
\(M=\frac{101}{204}< 1\left(đpcm\right)\)
Ta có: M=11.2.3 +12.3.4 +13.4.5 +...+1100.101.102
M=2.(11.2.3 +12.3.4 +13.4.5 +...+1100.101.102 ).12
M=(21.2.3 +22.3.4 +23.4.5 +...+2100.101.102 ).12
M=(11.2 -12.3 +12.3 -13.4 +13.4 -14.5 +...+1100.101 −1101.102 ).12
M=( 11.2 −1101.102 ).12
Mà 11.2 −1101.102 <1
Và 12 <1
=> (11.2 −1101.102 ) .12 <1
=> M <1
nhớ 9 k đó
Ta có: M=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\)
M=2.(\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +...+\(\frac{1}{100.101.102}\) ).\(\frac{1}{2}\)
M=(\(\frac{2}{1.2.3}\) +\(\frac{2}{2.3.4}\) +\(\frac{2}{3.4.5}\) +...+\(\frac{2}{100.101.102}\) ).\(\frac{1}{2}\)
M=(\(\frac{1}{1.2}\) -\(\frac{1}{2.3}\) +\(\frac{1}{2.3}\) -\(\frac{1}{3.4}\) +\(\frac{1}{3.4}\) -\(\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\) ).\(\frac{1}{2}\)
M=( \(\frac{1}{1.2}-\frac{1}{101.102}\)).\(\frac{1}{2}\)
Mà \(\frac{1}{1.2}-\frac{1}{101.102}<1\)
Và \(\frac{1}{2}<1\)
\(=>\) (\(\frac{1}{1.2}-\frac{1}{101.102}\) ) .\(\frac{1}{2}\) \(<1\)
\(=>\) M <1
giong nhu dap an minh viet khi nay do
nho k cho minh voi nha
A=1/2(2/1.2.3+2/2.3.4+...+2/2014.2015.2016)~A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+...+1/2014.2015-1/2015.2016)~~A=1/2(1/1.2-1/2015.2016)~A=1/2(1/2-1/4062240)~A=1/2.2031119/4062240~A=203119/8124480. Dấu/= dấu gạch ps còn ~ là dấu xuống dòng. Còn bài này thì ko biết dung hay sai nua
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)
\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2014.2015}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4058210}\right)\)
\(S=\frac{1}{2}.\left(\frac{2029105}{4058210}-\frac{1}{4058210}\right)\)
\(S=\frac{1}{2}.\frac{2029104}{4058210}\)
\(S=\frac{1014552}{4058210}\)
Chúc bạn học tốt !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
Ta có :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)
\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{2015.2016}\right):2\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016}\)
\(\Rightarrow A< \frac{1}{4}\)
Vậy A < \(\frac{1}{4}\)
_Chúc bạn học tốt_
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2014+2015+2016}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{2014.2015.2016}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)
\(\Rightarrow2A< \frac{1}{1.2}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{4}\)
Vậy ....
A= 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + ....... + 1/2014 - 1/2015 - 1/2016
Rồi đoạn sau tự tính tiếp nhé :)) Đến đôạn này chắc trừ được
s=1/1*2-1/2*3+1/2*3-1/3*4+....+1/2009*2010-1/210*2011
=1/1*2-1/2010*2011
<1/1*2
\(S=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2009\cdot2010\cdot2011}\)
\(S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}-\frac{1}{2010\cdot2011}\)
\(S=\frac{1}{1\cdot2}-\frac{1}{2010\cdot2011}\)
\(S=\frac{1}{2}-\frac{1}{2010\cdot2011}< \frac{1}{2}\)
=> S < P
\(M=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{100\cdot101\cdot102}\\ M=\frac{1}{2}\cdot\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{100\cdot101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\left(\frac{5151}{10302}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\frac{25}{51}\\ M=\frac{25}{102}\\ \Rightarrow M< 1\)
Vậy M < 1
M<1 ok?