A=4^2015+4^2016+4^2017+4^2018. chung minh rang tong A chia hết cho 5
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TP
1
11 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1+\frac{3}{2015}\right)\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(=\left(1+1+1+1\right)+\left(\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\right)\)
\(=4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\)
Vì \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2015}>\frac{1}{2017};\frac{1}{2015}>\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2015}-\frac{1}{2016}>0;\frac{1}{2015}-\frac{1}{2017}>0;\frac{1}{2015}-\frac{1}{2018}>0\)
\(\Rightarrow\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>0\)
\(\Rightarrow4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>4\)
\(\Rightarrow A>4\left(dpcm\right)\)
LG
0
24 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)
Xét :
\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)
Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm )
...
VN
3
20 tháng 8 2020
Đặt \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
Ta có: \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
\(\Leftrightarrow4A_1=4+4^2+4^3+...+4^{2017}+4^{2018}\)
Lấy \(4A_1-A_1\)ta có:
\(4A_1-A_1=\left(4+4^2+4^3+...+4^{2017}+4^{2018}\right)-\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
\(\Leftrightarrow3A_1=4^{2018}-1\)
\(\Leftrightarrow A_1=\frac{4^{2018}-1}{3}\)
Thay \(A_1=\frac{4^{2018}-1}{3}\)vào biểu thức A, ta có:
\(A=75.\left(\frac{4^{2018}-1}{3}\right)+25\)
\(\Leftrightarrow A=25.\left(4^{2018}-1\right)+25\)
\(\Leftrightarrow A=25.4^{2018}⋮4^{2018}\)
Vậy \(A⋮4^{2018}\)
chúc bn hok tốt
TT
1
YY
19 tháng 4 2018
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\frac{2016-1}{2016}+\frac{2017-1}{2017}+\frac{2018-1}{2018}+\frac{2015+3}{2015}\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(=4+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2015}-\frac{1}{2018}\)
mà \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2017};\frac{1}{2018}\)
\(\Rightarrow A>4\)
NT
0
A=4^2015+4^2016+4^2017+4^2018
A=(4^2015+4^2016)+(4^2017+4^2018)
A=4^2015.(1+4)+4^2017.(1+4)
A=5.(4^2015+4^2017)
=>A chia hết cho 5