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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 ~
A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100
Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100
A= 99/100
Cảm ơn bạn Kudo Shinichi, nhưng
1=2-1 ->ok
1/2=1-1/2 ->ok
1/3=1/2-1/3 -> sai
vì 1/2-1/3=1/6
@@@) Ta có: \(A=\frac{5^{2016}+4}{5^{2015}+4}\Rightarrow\frac{1}{5}A=\frac{5^{2016}+4}{5^{2016}+20}=1+\frac{-16}{5^{2016}+20}\)
\(B=\frac{5^{2014}+4}{5^{2013}+4}\Rightarrow\frac{1}{5}B=\frac{5^{2014}+4}{5^{2014}+20}=1+\frac{-16}{5^{2014}+20}\)
Ta thấy: \(1+\frac{-16}{5^{2016}+20}>1+\frac{-16}{5^{2014}+20}\) =>\(\frac{1}{5}A>\frac{1}{5}B\Rightarrow A>B\)
Bài thứ 2 sai để nhé hai cái đó = nhau mà
\(A=5+5^2+5^3+5^4+...+5^{2004}\)
\(5A=5^2+5^3+5^4+5^5+...+5^{2005}\)
\(5A-A=\left(5^2+5^3+5^4+5^5+...+5^{2005}\right)-\left(5+5^2+5^3+5^4+...+5^{2004}\right)\)
\(4A=5^{2005}-5\)
\(A=\dfrac{5^{2005}-5}{4}\)
\(B=7^1+7^2+7^3+....+7^{2015}\)
\(7B=7^2+7^3+7^4+....+7^{2016}\)
\(7B-B=\left(7^2+7^3+7^4+...+7^{2016}\right)-\left(7+7^2+7^3+....+7^{2015}\right)\)
\(6B=7^{2016}-7\)
\(B=\dfrac{7^{2016}-7}{6}\)
\(C=4^5+4^6+4^7+...+4^{2016}\)
\(4C=4^6+4^7+4^8+...+4^{2017}\)
\(4C-C=\left(4^6+4^7+4^8+...+4^{2017}\right)-\left(4^5+4^6+4^7+...+4^{2016}\right)\)
\(3C=4^{2017}-4^5\)
\(C=\dfrac{4^{2017}-4^5}{3}\)
A = 5 + 52 + 53 + 54 + ... + 52004
5A = 52 + 53 + 54 + 55 + ... + 52005
5A - A = 52005 - 5
4A = 52005 - 5
A = (52005 - 5) : 4
B = 71 + 72 + 73 + ... + 72015
7B = 72 + 73 + 74 + ... + 72016
7B - B = 72016 - 7
6B = 72016 - 7
B = (72016 - 7) : 6
C = 45 + 46 + 47 + ... + 42016
4C = 46 + 47 + 48 + ... + 42017
4C - C = 42017 - 45
3C = 42017 - 45
C = (42017 - 45) : 3
\(A=1+7+7^2+7^3+...+7^{2016}\)
\(\Rightarrow7A=7\left(1+7+7^2+7^3+...+7^{2016}\right)\)
\(7A=7+7^2+7^3+7^4+...+7^{2017}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2017}\right)-\left(1+7+7^2+...+7^{2016}\right)\)
\(\Rightarrow6A=7^{2017}-1\)
\(\Rightarrow A=\dfrac{7^{2017}-1}{6}\)
a, \(5S=5^2+5^3+...+5^{2017}\)
\(5S-S=5^{2017}-5\)
\(S=\frac{5^{2017}-5}{4}\)
b,\(3S=3^2+3^3+...+3^{101}\)
\(3S-S=3^{101}-3\)
\(S=\frac{3^{101}-3}{2}\)
c, \(3S=3-3^2+3^3-...-3^{2016}\)
\(3S+S=1-3^{2016}\)
\(4S=1-3^{2016}\)
\(S=\frac{1-3^{2016}}{4}\)
b, 3S = 3^2+3^3+.....+3^101
2S=3S-S=(3^3+3^3+.....+3^101)-(3+3^2+....+3^100) = 3^101-3
=> S = (3^101-3)/2
Tk mk nha