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\(M=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{100\cdot101\cdot102}\\ M=\frac{1}{2}\cdot\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{100\cdot101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{101\cdot102}\right)\\ M=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\left(\frac{5151}{10302}-\frac{1}{10302}\right)\\ M=\frac{1}{2}\cdot\frac{25}{51}\\ M=\frac{25}{102}\\ \Rightarrow M< 1\)
Vậy M < 1
\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\right)\)
\(M=\frac{1}{2}.\left(1-\frac{1}{102}\right)\)
\(M=\frac{101}{204}< 1\left(đpcm\right)\)
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
\(B=-\frac{3}{5}\left(\frac{3}{8}-2+\frac{5}{8}\right)\)
\(B=-\frac{3}{5}.\left(-1\right)=\frac{3}{5}\)
\(C=\frac{8}{5}.\frac{3}{4}-\left(\frac{11}{20}-\frac{1}{4}\right).\frac{7}{3}\)
\(C=\frac{6}{5}-\frac{3}{10}.\frac{7}{3}\)
\(C=\frac{6}{5}-\frac{7}{10}=\frac{1}{2}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}<\frac{1}{4}\)
B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}< 3\)
A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)
= 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)
= 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\)) - \(\frac{1}{21}\)
= 1 - \(\frac{1}{21}\)
= \(\frac{20}{21}\)< \(\frac{1}{4}\)
=> Đề bài có sai ko bạn?
bạn phải cho ra 2 số cuối thì mới làm đc nha có 1 s
ố cuối ko làm đc đâu
A= 1-1/2 + 1-1/3 + 1/2-1/5 + 1/3-1/8+ 1/5-1/13+1/8- 1/21 +....+ 1/610- 1/1597
A= 1/610
hello