chứng tỏ: 3/5.11 + 5/11.21 + 7/21.35 + 9/35.53 < 1
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PC
0
NH
0
MA
2
25 tháng 12 2021
Ta có \(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
\(\dfrac{1}{5^3}< \dfrac{1}{4.5.6}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)\\ ...\\ \dfrac{1}{2021^3}< \dfrac{1}{2020.2021.2022}=\dfrac{1}{2}\left(\dfrac{1}{2020.2021}-\dfrac{1}{2021.2022}\right)\)
Cộng VTV ta được
\(VT< \dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2021.2022}\right)=\dfrac{1}{12}-\dfrac{1}{2\left(2021.2022\right)}< \dfrac{1}{12}\)
V2
25 tháng 12 2021
\(n^3=n.n^2>n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}.\dfrac{n+1-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)
\(\dfrac{1}{3^3}+\dfrac{1}{5^3}+.......+\dfrac{1}{2009^3}< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+.....\dfrac{1}{2008.2009.2010}=\dfrac{1}{2}\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.........+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2009.2010}\right)\)
\(=\dfrac{1}{2}\)
5 tháng 9 2015
73=343 đồng dư với 1(mod 9)
=>(73)6=718 đồng dư với 1(mod 9)
=>718=9k+1
=>B=9k+1+18.3-1=9k+18.3=9(k+2.3) chia hết cho 9
=>đpcm
KT
3
25 tháng 8 2021
Xét vế phải :
\(VT=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{10}-\frac{1}{106}