Cho T= 2/21+3/22+4/23+...+2017/22016
So sánh T với 3
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AT
11 tháng 7 2017
1, \(A=2.3^4+2^3=2\left(3^4+2^2\right)=2.85=170\)
2,\(=>9A=3^{13}+3^{15}+3^{17}+...+3^{25}\)
\(=>9A-A=3^{25}-3^{11}\)
\(=>A=\dfrac{3^{25}-3^{11}}{8}\)
Ta thấy : \(3^{25}=3.3^{4.6}=3\times.........1=...........3\)
Lại có: \(3^{11}=3^3.3^{4.2}=27\times.........1=.......7\)
=> \(=>3^{25}-3^{11}=....3-......7=.....6\)
Ta có: \(A=\dfrac{.............6}{8}=>A=.........2;A=.....7\)
Mà số chia hết cho 5 có tận cùng là 0 ; 5 nên => A không chia hết cho 5;
3,\(B=\dfrac{2017^{17}\left(2017^{2000}-1\right)}{2017^{2016}.2017^{2002}}\)
\(=>B=\dfrac{2017^{2000}-1}{2017^{2001}}\)
CHÚC BẠN HK TỐT....
DB
0
LK
2
27 tháng 6 2021
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
PL
1
K
24 tháng 6 2015
Vì: \(\frac{3}{21}=\frac{3}{21}\)
\(\frac{3}{22}\) < \(\frac{3}{21}\)
\(\frac{3}{23}\) < \(\frac{3}{21}\)
\(\frac{3}{24}\)<\(\frac{3}{21}\)
\(\frac{3}{25}\)< \(\frac{3}{21}\)
.....
\(\frac{2}{29}\)<\(\frac{3}{21}\)
\(\frac{2}{30}\)<\(\frac{3}{21}\)
Nên \(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{21}.10\)
Ta có: \(\frac{3}{21}.10\) = \(\frac{10}{7}\)
Mà \(\frac{10}{7}\) < \(\frac{3}{2}\)
=>\(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{2}\)
Vậy E < M
8 tháng 10 2018
Ta có :
\(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2015+2\sqrt{2015.2017}+2017=8064+2\sqrt{2015.2017}\)
\(\left(2\sqrt{2016}\right)^2=8064\)
Vì \(\left(\sqrt{2015}+\sqrt{2017}\right)^2>\left(2\sqrt{2016}\right)^2\) nên \(\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)
Vậy...
Chúc bạn học tốt ~
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\)
\(\frac{1}{2}T=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
\(T-\frac{1}{2}T=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
\(\frac{1}{2}T=1+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-...-\frac{2017}{2^{2017}}\)
\(\frac{1}{2}T=1+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
\(\frac{1}{2}T=1+\left(\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{2^{2016}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{2016}}\)
Mà \(\frac{1}{2^{2016}}>0\)
\(\Rightarrow\)\(A=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
\(\Leftrightarrow\)\(1+A-\frac{2017}{2^{2017}}< 1+\frac{1}{2}-\frac{1}{2^{2016}}-\frac{2017}{2^{2017}}\)
\(\Leftrightarrow\)\(\frac{1}{2}T< \frac{3}{2}-\left(\frac{1}{2^{2016}}+\frac{2017}{2^{2017}}\right)\)
Mà \(\frac{1}{2^{2016}}+\frac{2017}{2^{2017}}\)
\(\Rightarrow\)\(\frac{1}{2}T< \frac{3}{2}\)
\(\Rightarrow\)\(T< \frac{3}{2}.2\)
\(\Rightarrow\)\(T< 3\)
Vậy \(T< 3\)
Chúc bạn học tốt ~
\(T< 3\)