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Ta có:5/20>5/25
5/21>5/25
5/22>5/25
5/23>5/25
5/24>5/25
=>S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=1
=>5/20+5/21+5/22+5/23+5/24>1
DỄ
DO: 5/20 <1
5/21<1
5/22<1
5/23<1
5/24<1
=> 5/20+5/21+5/22+5/23+5/24<1
hay S<1 ( ĐPCM)
ĐÚNG NÈ ỦNG HỘ
Ta có:\(\dfrac{1}{20}>\dfrac{1}{21}>\dfrac{1}{22}>\dfrac{1}{23}>\dfrac{1}{24}>\dfrac{1}{25}\)
=>S=\(\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}=5\left(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{24}\right)>5\cdot\left(\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{25}\right)\)
=>S>\(5\cdot\dfrac{5}{25}\)
=>S>1(đpcm)
Sửa đề : Chứng minh : S > 1
Ta thấy : \(\frac{5}{20}>\frac{5}{21}>\frac{5}{22}>\frac{5}{23}>\frac{5}{24}\)
\(\Rightarrow S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>\frac{5}{24}\times5=\frac{25}{24}>1\)
Vậy S > 1 (ĐPCM)
\(S=\frac{2016}{2.3:2}+\frac{2016}{3.4:2}+...+\frac{2016}{2015.2016:2}\)
\(S=\frac{4032}{2.3}+\frac{4032}{3.4}+...+\frac{4032}{2015.2016}\)
\(S=4032\left[\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right]\)
\(S=4032\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right]\)
\(S=4032\left[\frac{1}{2}-\frac{1}{2016}\right]=4032\cdot\frac{1007}{2016}\)
\(S=2014\)
S = \(2016+\frac{2016}{1+2}+\frac{2016}{1+2+3+}+...+\frac{2016}{1+2+3+...+2015}\)
S = \(2016+\left(\frac{2016}{1+2}+\frac{2016}{1+2+3}+...+\frac{2016}{1+2+3+...+2015}\right)\)
S = \(2016+2016.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\right)\)
đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\)
A = \(\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2015\right).2015:2}\)
A = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2015.2016}\)
A = \(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+...+2.\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
A = \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
A = \(2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
A = \(2.\frac{1007}{2016}=\frac{1007}{1008}\)
Thay A vào ta được :
S = \(2016+2016.\frac{1007}{1008}\)
S = \(2016.\left(1+\frac{1007}{1008}\right)\)
S = \(2016.\frac{2015}{1008}\)
S = \(4030\)
98
SSH của S là : ( 2016 - 20 ) : 2 + 1 = 999 ( số hạng )
<=> S = ( 2016 + 20 ) . 999 : 2 = 1016982
Vậy S = 1016982