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M=2018^2-2017^2+2016^2-2015^2+............+2^2-1^2
M=(2018+2017).(2018-2017)+(2016+2015).(2016-2015)+...........+(2+1).(2-1)
M=2018+2017+2016+2015+.................+2+1
M=2018.(2018+1)/2=2018.2019/2
M=1009.2019M=2037171
(1/2+2015/2016+1 )nhaân (2016/2017+7/2) - (1/2+2015/2016) nhaân (7/2+2016/2017 +1)
TÍNH PHEP TÍNH NAY
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\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)
\(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)
\(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Vậy \(A=\frac{2^{2018}-1}{2^{2017}}\)
A=đã cho.
2A=1+1/2+1/2^2+1/2^3+...+1/2^2016.
2A-A=1-1/2^2017(khử).
A=1-1/2^2017.
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=1/2 - 1/3 +1/3 - 1/4 +...+1/2016 - 1/2017
=1/2 - 1/2017
=...
1/2:3+1/3:4+1/4:5+...1/2016:2017
1/2.1/3+1/3.1/4+1/4.1/5+...1/2016.1/2017
1/2.3+1/3.4+1/4.5+...1/2016.2017
1/2-1/3+1/3-1/4+1/4-1/5+...1/2016-1/2017
=1/2-1/2017
=2017/4034-2/4034
=2015/4034
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999 - 888 - 111 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111 + 111 - 111
= 0 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111
= 0 + 111 - 111
= 111 - 111
= 0