a, \(S=1+3+3^2+...+3^{2019}\)

\(3S=3+3^2+3^3+...+3^{2020}\)

\(3S-S=\left(3+3^2+3^3+...+3^{2020}\right)-\left(1+3+3^2+...+3^{2019}\right)\)

\(2S=3^{2020}-1\)

\(S=\frac{3^{2020}-1}{2}\)

b, \(S=1+3+3^2+3^3+...+3^{2019}\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\)

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

\(S=4\cdot1+3^2\cdot4+...+3^{2018}\cdot4\)

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

\(7^{1999}=7^{1996}\cdot7^3=\overline{......1}\cdot\overline{......3}=\overline{......3}\)

\(8^{2015}=8^{2012}\cdot8^3=\overline{.....6}\cdot\overline{......2}=\overline{......2}\)

\(9^{3^2}=9^9=9^8\cdot9=\overline{......1}\cdot\overline{......9}=\overline{.....9}\)

\(87^{32}=\overline{......1}\)

\(58^{33}=58^{32}\cdot58=\overline{.....6}\cdot58=\overline{.....8}\)

2^500 = (2^5)^100 = 32^100

5^200 = (5^2)^100 = 25^100

=> 2^500 > 5^200

3^200 = (3^2)^100 = 9^100

2^300 = (2^3)^100 = 8^100

=> 3^200 > 2^300

Ta có : 2⁵⁰⁰ =(2⁵)¹⁰⁰ =32¹⁰⁰

           5²⁰⁰ =(5²)¹⁰⁰ =25¹⁰⁰

So sánh : 32¹⁰⁰< 25¹⁰⁰

Nên :         2⁵⁰⁰<5²⁰⁰

Câu kia cũng tương tự nhé =))

\(2A=2+2^2+2^3+...+2^{2009}\)

\(A=2A-A=2^{2009}-1=B\)

2A=2+22+23+...+22009

A=2A-A=2^{2009}-1=BA=2A−A=22009−1=B

Help!!

Ủa bạn,có bị lộn đề ko zị

thanks you nhiều

( x + 1) + (x + 2) + (x+3) +…+ (x + 100) = 7450

=> x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450

=> (x + x + x + .... + x) + (1 + 2 + 3 + ... + 100) = 7450

=> 100x + 5050 = 7450

=> 100x = 1400

=> x = 14

\(2015^2.\left(x-10\right)=2015^6\)

\(\Rightarrow x-10=2015^6:2015^2\)

\(\Rightarrow x-10=2015^4\)

Từ đây tự tính tiếp nhé em!

2015⁵ × ( x - 10 ) = 2015⁶ 

               ( x - 10 ) = 2015⁶ ÷ 2015⁵

                 x - 10    = 2015

                       x      = 2015 + 10

                       x      = 2025

Chúc bạn học tốt nha!