Câu 3:

\(A=3+3^2+...+3^{100}\)

\(3A=3^2+3^3+...+3^{101}\)

\(3A-A=3^2+3^3+...+3^{101}-\left(3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-3\) 

Mà: \(2A+3=3^N\)

\(\Rightarrow3^{101}-3+3=3^N\)

\(\Rightarrow3^{101}=3^N\)

\(\Rightarrow N=101\)

Vậy: ... 

Câu 1:

\(A=4+2^2+...+2^{20}\)

Đặt \(B=2^2+2^3+...+2^{20}\)

=>\(2B=2^3+2^4+...+2^{21}\)

=>\(2B-B=2^3+2^4+...+2^{21}-2^2-2^3-...-2^{20}\)

=>\(B=2^{21}-4\)

=>\(A=B+4=2^{21}-4+4=2^{21}\) là lũy thừa của 2

Câu 6:

Đặt A=1+2+3+...+n

Số số hạng là \(\dfrac{n-1}{1}+1=n-1+1=n\left(số\right)\)

=>\(A=\dfrac{n\left(n+1\right)}{2}\)

=>\(A⋮n+1\)

Câu 5:

\(A=5+5^2+...+5^8\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)

\(=30\left(1+5^2+5^4+5^6\right)⋮30\)

`A=1+2+2^2+2^3+2^4+...+2^{200}`

`=>2A=2+2^2+2^3+2^4+2^5+...+2^{201}`

`=>2A-A=(2+2^2+2^3+2^4+2^5+...+2^{201})-(1+2+2^2+2^3+2^4+...+2^{200})`

`=>A=2^{201}-1`

`=>A+1=2^{201}`

A=2+22+23+...+220A=2+22+23+...+220

2A=22+23+24+...+2212A=22+23+24+...+221

2A−A=(22+23+24+...+221)−(2+22+23+...+220)2A−A=(22+23+24+...+221)−(2+22+23+...+220)

A=221−2=24.5+1−2=(24)5.2−2=165.2−2A=221−2=24.5+1−2=(24)5.2−2=165.2−2

A=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯.......6.2−2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯........2−2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯...........0A=.......6¯.2−2=........2¯−2=...........0¯

Vậy chữ số tận cùng cả A là 0

a bang 2

b i don't know

a/\(A=1+2+2^2+.......+2^{200}\)

\(\Leftrightarrow2A=2+2^2+..........+2^{200}+2^{201}\)

\(\Leftrightarrow2A-A=\left(2+2^2+..........+2^{201}\right)-\left(1+2+.....+2^{200}\right)\)

\(\Leftrightarrow A=2^{201}-1\)

a) \(A=1+2+2^2+2^3+...+2^{99}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{100}-1-2-2^2-...-2^{99}=2^{100}-1\)

b) \(A=1+2+2^2+...+2^{99}=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)

\(=15+2^4.15+...+2^{96}.15=15\left(1+2^4+...+2^{96}\right)\)

\(=3.5\left(1+2^4+...2^{96}\right)\) chia hết cho 3 và 5

c) \(A=1+2+2^2+...+2^{99}\)

\(=1+2\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)

\(=1+2.7+...+2^{97}.7=1+7\left(2+...+2^{97}\right)\) chia 7 dư 1

=> A không chia hết cho 7

Đổi 4 thành 2 mũ 2

Thử xem cs đúng ko . Vì mik chữ thầy toán giả thầy toán hết r

Dễ:đổi 4=2²

B=2²+2³+2⁴+...+2²⁰

ta có:B=2B-B=(2³+2⁴+2⁵+...+2²¹)-(2²+2³+2⁴+...+2²⁰)

                    = 2²¹-2²

Nói trước: đây là mình rút gọn chứ viết mà theo cơ số 2 thì khó quá

0,08%=0,0008

Diện tích thực tế là 32:0,01%=320000cm²

1, 

\(64^7\div4^5\)

\(=\left(4^3\right)^7\div4^5\)

\(=4^{21}\div4^5\)

\(=4^{16}\)

2, 

\(A=2+2^2+2^3+...+2^{2019}\)

\(2A=2^2+2^3+2^4+...+2^{2020}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2020}\right)-\left(2+2^2+2^3+...+2^{2019}\right)\)

\(A=2^{2020}-2\)

3, 

\(74^{30}=\left(74^2\right)^{15}=\overline{.....6}^{15}=\overline{.....6}\)

\(39^{31}=39^{30}\cdot39=\left(39^2\right)^{15}\cdot39=\overline{.....1}^{15}\cdot39=\overline{.....1}\cdot39=\overline{......9}\)

\(87^{32}=\left(87^4\right)^8=\overline{.....1}^8=\overline{.....1}\)

\(58^{33}=58^{32}\cdot58=\left(58^4\right)^8\cdot58=\overline{....6}^8\cdot58=\overline{.....6}\cdot58=\overline{....8}\)

\(23^{35}=23^{32}\cdot23^3=\left(23^4\right)^8\cdot\overline{....7}=\overline{....1}^8\cdot\overline{...7}=\overline{....1}\cdot\overline{....7}=\overline{....7}\)