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Với n là số lẻ thì n + 20172018 là số chẵn
Suy ra .............
Với n là số chẵn thì n + 20182017 là số chẵn
Suy ra ............
Vậy ..............
a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)
Đặt \(B=2+2^2+...+2^{2017}\)
\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)
\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)
\(\Rightarrow B=2^{2018}-2\)
\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)
\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)
\(\Rightarrow A=-2\)
b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)
\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)
\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)
Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
Ta có:
\(A=2017^{10}=2017^8.2017^2\)
Lại có:
\(2017^8.2017^2< 2017^8.2018^2\)
\(\Rightarrow A< B\)
Ta có
\(2017^{2017}=\left(2017^{2016}\right).2017=\left(...1\right).2017=\left(...7\right)\)
\(2013^{2013}=\left(2013^{2012}\right).2013=\left(...1\right).2013=\left(...3\right)\)
\(\Rightarrow2013^{2013}+2017^{2017}=\left(...3\right)+\left(...7\right)=\left(...0\right)⋮10\)
\(2013^{2013}+2017^{2017}\)
Ta có:
\(2013^{2013}=\left(2013^{2012}\right).2013=\overline{...1}.2013=\overline{...3}\)
\(2017^{2017}=\left(2017^{2016}\right).2017=\overline{...1}.2017=\overline{...3}\)
\(\Rightarrow2013^{2013}+2017^{2017}=\overline{...3}+\overline{...7}=\overline{...0}⋮10\)
\(\Rightarrow2013^{2013}+2017^{2017}⋮10\)
ta có:
\(B=\dfrac{2016}{2017}+\dfrac{2017}{2018}>\dfrac{2016}{2018}+\dfrac{2017}{2018}=\dfrac{2016+2017}{2018}>\dfrac{2016+2017}{2017+2018}=A\)
Có : S = (2017+2017^2)+(2017^3+2017^4)+.....+(2017^9+2017^10)
= 2017.(1+2017)+2017^3.(1+2017)+......+2017^9.(1+2017)
= 2017.2018+2017^3.2018+......+2017^9.2018
= 2018.(2017+2017^3+....+2017^9) chia hết cho 2018
Tk mk nha
Dãy số trên có 10 số hạng chia thành 5 nhóm mỗi nhóm có 2 số hạng
Ta có:
S=(2017+2017^2)+(2017^3+2017^4)+..........+(2017^9+2017^10)
S=(2017.1+2017.2017)+.........+(2017^9.1+2017^9.2017)
S=2017.(2017+1)+.....+2017^9.(2017+1)
S=2017.2018+......+2017^9.2018
S=2018.(2017+.....+2017^9)
=>S chia hết chp 2018
k cho tớ nha!!!!!