a, S = 1 + 2 + 2² + 2³ + ... + 2²⁰¹⁷

Ta có : 2S = 2 + 2² + 2³ +.... + 2²⁰¹⁸

Lấy 2S - S ta được : S = 2²⁰¹⁸ - 1

b, Đặt S = 3 + 3² + 3³ + ... + 3²⁰¹⁷

Ta có : 3S = 3² + 3³ + ... + 3²⁰¹⁸

Lấy 3S - S ta được 2S = 3²⁰¹⁸ -3 

=> \(S=\frac{3^{2018}-3}{2}\)

c, Đặt S = 4 + 4² + 4³ + ... + 4²⁰¹⁷ 

Ta có : 4S = 4² + 4³ + ... + 4²⁰¹⁸

Lấy 4S - S ta được 3S = 4²⁰¹⁸ - 4 

=> \(S=\frac{4^{2018}-4}{3}\)

a, S = 1 + 2 + 22 + 23 + ... + 22017

Ta có : 2S = 2 + 22 + 23 +.... + 22018

Lấy 2S - S ta được : S = 22018 - 1

b, Đặt S = 3 + 32 + 33 + ... + 32017

Ta có : 3S = 32 + 33 + ... + 32018

Lấy 3S - S ta được 2S = 32018 -3 

=> �=32018−32S=232018−3​

c, Đặt S = 4 + 42 + 43 + ... + 42017 

Ta có : 4S = 42 + 43 + ... + 42018

Lấy 4S - S ta được 3S = 42018 - 4 

=> �=42018−43S=342018−4​
 

a) - 10 b) - 1009

\(B=1+1^2+1^3+.......+1^{2017}\)

\(1.B=1^2+1^3+....+1^{2018}\)

\(1B-B=1^{2018}-1\)

\(B.0=1^{2018}-1\)

\(B=2+2^2+2^3+.....+2^{2017}\)

\(2B=2^2+2^4+.....+2^{2018}\)

\(2B-B=2^{2018}-2\)

\(B=\frac{2^{2018}-2}{1}\)

\(B=3+3^2+3^3+.....+3^{2017}\)

\(3B=3^2+3^3+....+3^{2018}\)

\(3B-B=2B=3^{2018}-3\)

\(B=\frac{3^{2018}-3}{2}\)

Nhớ k cho mình nhé! Thank you!!!

Ta có: \(\frac{3}{1\times2}+\frac{3}{2\times3}+\cdots+\frac{3}{2016\times2017}\)

\(=3\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2016\times2017}\right)\)

\(=3\times\left(1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=3\times\left(1-\frac{1}{2017}\right)=3\times\frac{2016}{2017}=\frac{6048}{2017}\)

=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.......+\(\frac{1}{2016}\)-\(\frac{1}{2017}\)+1

=\(\frac{1}{1}\)-\(\frac{1}{2017}\)+1

=\(\frac{2016}{2017}\)+1

=\(\frac{1}{2017}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}+1\)

\(=1-\frac{1}{2017}+1\)

\(=\frac{2016}{2017}+1\)

\(=\frac{4033}{2017}\)

\(a)\) \(S=1+2+2^2+2^3+...+2^{2017}\)

\(2S=2+2^2+2^3+2^4+...+2^{2018}\)

\(2S-S=\left(2+2^2+2^3+2^4+...+2^{2018}\right)-\left(1+2+2^2+2^3+...+2^{2017}\right)\)

\(S=2^{2018}-1\)

\(b)\) \(S=3+3^2+3^3+...+3^{2017}\)

\(3S=3^2+3^3+3^4+...+3^{2018}\)

\(3S-S=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)

\(2S=3^{2018}-3\)

\(S=\frac{3^{2018}-3}{2}\)

\(c)\) \(S=4+4^2+4^3+...+4^{2017}\)

\(4S=4^2+4^3+4^4+...+4^{2018}\)

\(4S-S=\left(4^2+4^3+4^4+...+4^{2018}\right)-\left(4+4^2+4^3+...+4^{2017}\right)\)

\(3S=4^{2018}-4\)

\(S=\frac{4^{2018}-4}{3}\)

\(d)\) \(S=5+5^2+5^3+...+5^{2017}\)

\(5S=5^2+5^3+5^4+...+5^{2018}\)

\(5S-S=\left(5^2+5^3+5^4+...+5^{2018}\right)-\left(5+5^2+5^3+...+5^{2017}\right)\)

\(4S=5^{2018}-5\)

\(S=\frac{5^{2018}-5}{2}\)

Chúc em học tốt ~ 

Tks anh ạ 

=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

vip

vip

vip

chúc bạn học ngu

bai nay mk hoc tu lop 5

\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)

\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)

\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)

\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)

\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)

Vậy \(S=\frac{1008}{1009}\)

\(S=\frac{1008}{1009}\)