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Xét thấy vế trái có \(\left(2017-1\right):1+1=2017\)số hạng
Do đó có thể biến đổi \(VT=-\left(1\cdot2\cdot3\cdot...\cdot2017\right)< 0\forall x\)
Vậy...
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a, \(S=1+2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)
\(\Rightarrow S=2^{2018}-1\)
Vậy : \(S=2^{2018}-1\)
b, Ta có : \(2^{2018}-1< 2^{2018}=2^2.2^{2016}=4.2^{2016}< 5.2^{2016}\)
Vì : \(2^{2018}-1< 4.2^{2016}< 5.2^{2016}\Rightarrow S< 5.2^{2016}\)
Vậy : \(S< 5.2^{2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+2+...+2^{2017}\)
\(2S=2+2^2+...+2^{2018}\)
\(2S-S=2+2^2+...+2^{2018}-1-2-...-2^{2017}\)
\(S=2^{2018}-1\)
\(S=3+3^2+...+3^{2017}\)
\(3S=3^2+3^3+...+3^{2018}\)
\(3S-S=3^2+3^3+...+3^{2018}-3-3^2-...-3^{2017}\)
\(2S=3^{2018}-3\)
\(S=\dfrac{3^{2018}-3}{2}\)
\(S=4+4^2+...+4^{2017}\)
\(4S=4^2+4^3+...+4^{2018}\)
\(4S-S=4^2+4^3+...+4^{2018}-4-4^2-...-4^{2017}\)
\(3S=4^{2018}-4\)
\(S=\dfrac{4^{2018}-4}{3}\)
\(S=5+5^2+...+5^{2017}\)
\(5S=5^2+5^3+...+5^{2018}\)
\(5S-S=5^2+5^3+...+5^{2018}-5-5^2-...-5^{2017}\)
\(4S=5^{2018}-5\)
\(S=\dfrac{5^{2018}-5}{4}\)
a) S=1+2+22+...+22017
=> 2S=2.(1+2+22+...+22017)
=>2S=2+22+23+...+22018
=>S=(2+22+23+ ..+22018) - (1+2+22+ ....+22017 )
=> S =22018-1
![](https://rs.olm.vn/images/avt/0.png?1311)
So sánh S với 2 biết :
S = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2017}{2^{2017}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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!!!!!