Cho : \(S=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+...+\dfrac{6026}{2008.2009.2010}\). So sánh S với 2
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Cho : \(S=\frac{5}{1.2.3}+\frac{8}{2.3.4}+...+\frac{6026}{2008.2009.2010}\) . So sánh S với 2
Xét tổng gồm 2008 số hạng S=5/1.2.3+8/2.3.4+...+6026/2008.2009.2010
So sánh S với 2
Xét tổng gồm 2008 số hạng \(S=\frac{5}{1.2.3}+\frac{8}{2.3.4}+...+\frac{6026}{2008.2009.2010}.\)So sánh S với 2
Cho S=\(\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+...+\dfrac{6068}{2022.2023.2024}\)
So sánh S với 2
Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2}{n\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}+\frac{1}{n+1}-\frac{1}{n+2}\)
\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)
\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)
...
Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)
=>\(S_{2022}=\frac{5}{1\cdot2\cdot3}+\frac{8}{2\cdot3\cdot4}+\cdots+\frac{6068}{2022\cdot2023\cdot2024}<2\)
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Xét tổng gồm 2008 số hạng S= 5/(1.2.3) + 8/(2.3.4) + ......+ 6028/(2008.2009.2010).So sánh S với 2
Xét tổng gồm 2008 số hạng S= 5/(1.2.3) + 8/(2.3.4) +....+ 6028/(2008.2009.2010). So sánh S với 2?
11 tháng 4 2021 lúc 20:05
Cho :
\(A=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+...+\dfrac{6056}{2018.2019.2020}\)
Hãy so sánh A với 2
Đặt S=A
Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2}{n\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}+\frac{1}{n+1}-\frac{1}{n+2}\)
\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)
\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)
...
Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)
=>\(S_{2018}<2\)
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\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\)
\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\)