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\(a)\) \(S=1+2+2^2+2^3+...+2^{2017}\)
\(2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(2S-S=\left(2+2^2+2^3+2^4+...+2^{2018}\right)-\left(1+2+2^2+2^3+...+2^{2017}\right)\)
\(S=2^{2018}-1\)
\(b)\) \(S=3+3^2+3^3+...+3^{2017}\)
\(3S=3^2+3^3+3^4+...+3^{2018}\)
\(3S-S=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)
\(2S=3^{2018}-3\)
\(S=\frac{3^{2018}-3}{2}\)
\(c)\) \(S=4+4^2+4^3+...+4^{2017}\)
\(4S=4^2+4^3+4^4+...+4^{2018}\)
\(4S-S=\left(4^2+4^3+4^4+...+4^{2018}\right)-\left(4+4^2+4^3+...+4^{2017}\right)\)
\(3S=4^{2018}-4\)
\(S=\frac{4^{2018}-4}{3}\)
\(d)\) \(S=5+5^2+5^3+...+5^{2017}\)
\(5S=5^2+5^3+5^4+...+5^{2018}\)
\(5S-S=\left(5^2+5^3+5^4+...+5^{2018}\right)-\left(5+5^2+5^3+...+5^{2017}\right)\)
\(4S=5^{2018}-5\)
\(S=\frac{5^{2018}-5}{2}\)
Chúc em học tốt ~
\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2017}.\left(1+2+3+...+2017\right)\)
\(S=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{2017}.\frac{\left(1+2017\right).2017}{2}\)
\(S=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2018}{2}\)
\(S=\frac{1}{2}.\left(2+3+4+...+2018\right)\)
\(S=\frac{1}{2}.\frac{\left(2+2018\right).2017}{2}\)
\(S=\frac{2020.2017}{4}=505.2017=1018585\)
a) \(S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}\)
\(\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2\) vậy \(S=-2\)
Ta có :
\(S=\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{2016}{2017!}\)
\(S=\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{2017-1}{2017!}\)
\(S=\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+\frac{5}{5!}-\frac{1}{5!}+...+\frac{2017}{2017!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{2016!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{2017!}\)
\(S=\frac{1}{2}-\frac{1}{2017!}\)
Vậy \(S=\frac{1}{2}-\frac{1}{2017!}\)
Chúc bạn học tốt ~
\(S=1+5+5^2+5^3+.......+5^{2017}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+......+\left(5^{2016}+5^{2017}\right)\)
\(=6+5^2\left(1+5\right)+.........+5^{2016}\left(1+5\right)\)
\(=6+5^2.6+.......+5^{2016}.6=6\left(1+5^2+......+5^{2016}\right)⋮3\)
S=1+5+52+53+54+....+52017
S=(1+5)+(52+53)+(54+55)+.....+(52016+52017)
S=(1+5)+52.(1+5)+54.(1+5)+...+52016.(1+5)
S=6+52.6+54.6+...+52016.6
S=6.(1+52+54+...+52016)
S=2.3.(1+52+54+...+52016)\(⋮\)3
Chúc bn học tốt
Ta có :
\(S=5+5^2+5^3+...+5^{2016}+5^{2017}\)
\(=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2013}+5^{2014}+5^{2015}+5^{2016}\right)+5^{2017}\)
\(=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2012}\left(5+5^2+5^3+5^4\right)+5^{2017}\)
\(=\left(1+5^4+5^8+...+5^{2012}\right)\left(5+5^2+5^3+5^4\right)+5^{2017}\)
\(=\left(1+5^4+5^8+...+5^{2012}\right).65.12+5^{2017}\)
Ta có :
\(5^4\text{≡}1\left(mod13\right)\)
\(\Rightarrow\left(5^4\right)^{504}\text{≡}1^{504}\left(mod13\right)\)
\(\Rightarrow5^{2016}\text{≡}\left(mod13\right)\)
\(\Rightarrow5^{2017}\text{≡}5\left(mod13\right)\)
Lại có :
\(\left(1+5^4+5^8+...+5^{2012}\right).65.12\text{ }\text{⋮}65\)
\(5^{2017}\)không chia hết cho 65
\(\Rightarrow\left(1+5^4+5^8+...+5^{2012}\right).65.12+5^{2017}\)không chia hết cho 65
\(\Rightarrow S\)không chia hết cho 65
Vậy \(S\)không chia hết cho 65
\(S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2015}+5^{2016}\right)+5^{2017}\)
\(S=130+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+...+5^{2014}\left(5+5^2\right)+5^{2017}\)
\(S=130+5^2.130+5^4.130+...+5^{2014}.130+5^{2017}\)
\(S=130\left(1+5^2+5^4+...+5^{2014}\right)+5^{2017}\)
Vì \(S=130\left(1+5^2+5^4+...+5^{2014}\right)\)chia hết cho 65 nhưng \(5^{2017}\)không chia hết cho 65
=> \(S=130\left(1+5^2+5^4+...+5^{2014}\right)+5^{2017}\)không chia hết cho 65
Vậy \(5+5^2+5^3+5^4+5^5+...+5^{2017}\)Không chia hết cho 65
\(S=1+2+...+2^{2017}\)
\(2S=2+2^2+...+2^{2018}\)
\(2S-S=2+2^2+...+2^{2018}-1-2-...-2^{2017}\)
\(S=2^{2018}-1\)
\(S=3+3^2+...+3^{2017}\)
\(3S=3^2+3^3+...+3^{2018}\)
\(3S-S=3^2+3^3+...+3^{2018}-3-3^2-...-3^{2017}\)
\(2S=3^{2018}-3\)
\(S=\dfrac{3^{2018}-3}{2}\)
\(S=4+4^2+...+4^{2017}\)
\(4S=4^2+4^3+...+4^{2018}\)
\(4S-S=4^2+4^3+...+4^{2018}-4-4^2-...-4^{2017}\)
\(3S=4^{2018}-4\)
\(S=\dfrac{4^{2018}-4}{3}\)
\(S=5+5^2+...+5^{2017}\)
\(5S=5^2+5^3+...+5^{2018}\)
\(5S-S=5^2+5^3+...+5^{2018}-5-5^2-...-5^{2017}\)
\(4S=5^{2018}-5\)
\(S=\dfrac{5^{2018}-5}{4}\)
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