a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)

\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

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

\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)

\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)

Các ý dưới bạn làm tương tự nhé. 

*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)

              \(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)

              \(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)

              \(=6\times\left(2^2+2^3+...+2^{2008}\right)\)

              \(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)

               \(\Rightarrow A⋮3\)

*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)

               \(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)

               \(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)

               \(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)

                \(\Rightarrow A⋮7\)

Mình sửa lại đề C 1 chút xíu

*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)

               \(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)

               \(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)

                \(\Rightarrow C⋮4\)

Các câu khác làm tương tự nhé. Chúc bạn học tốt!

Thanks bạn

Giải: 

A= 2 + 2 mũ 2 + 2 mũ 3 + 2 mũ 4 +....+ 2 mũ 2010

A= (2 + 2 mũ 2) + (2 mũ 3 + 2 mũ 4) +....+ (2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 3) + 2 mũ 3 (1 + 2) + 2 mũ 2009 (1 +2_

A= 2.3 + 2 mũ 3.3 +....+ 2 mũ 2009.3

A= 3.(2 + 2 mũ 3 +....+ 2 mũ 2009) chia hết cho 3

A= (2 + 2 mũ 2 + 2 mũ 3) + (2 mũ 4 + 2 mũ 5 + 2 mũ 6) +....+ (2 mũ 2008 + 2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 2 + 2 mũ 2) + 2 mũ 4(1+ 2 + 2 mũ 2) +...+ 2 mũ 2008.(1 + 2 + 2 mũ 2)

A= 2.7 + 2 mũ 4. 7 +.... + 2 mũ 2008.7

A= 7.(2 + 2 mũ 4 +....+ 2 mũ 22010 chia hết cho 7.

Các câu còn lại làm tương tự như câu a nha bạn!

`A = 2 + 2^2+ ... + 2^2017`

`=> 2A = 2^2 + 2^3 + ... + 2^2018`

`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`

`=> A         = 2^2018 - 2`

`B = 1 + 3^2 + ... + 3^2018`

`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`

`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`

`=> 8B     = 3^2020 - 1`

`=> B       = (3^2020 - 1)/8`

`C = 5 + 5^2 - 5^3 + ... + 5^2018`

`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`

`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`

`=> 6C = 55 + 5^2019`

`=> C  = (5^2019 + 55)/6`

A=5(1+5^2)+5^5(1+5^2)+...+5^2021(1+5^2)

=26(5+5^5+...+5^2021) chia hết cho 26

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

a) \(S=1+5+5^2+5^3+...+5^{28}\)

\(S=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{27}+5^{28}\right)\)

\(S=1\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)

\(S=\left(1+5^2+...+5^{27}\right).6⋮3\left(dpcm\right)\)

b) \(S=1+5+5^2+5^3+...+5^{28}\)

\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{29}\)

\(\Rightarrow5S-S=\left(5+5^2+5^3+5^4+...+5^{29}\right)-\left(1+5+5^2+5^3+...+5^{28}\right)\)

\(\Rightarrow4S=5^{29}-1\)

\(\Rightarrow4S+1=5^{29}-1+1\)

\(\Rightarrow4S=5^{29}=5^n\)

\(\Rightarrow n=29\)

a) \(S=1+5+5^2+5^3+...+5^{28}\)

\(\Rightarrow S=\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)

\(\Rightarrow S=6+5^2.6+...+5^{27}.6\)

\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮6\)

\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮3\)

\(\Rightarrow dpcm\)

b) Bạn xem lại đề

1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) Bạn hãy xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2

Giải:

a) Đặt:

\(A=1+2^2+2^3+2^4+...+2^{2018}\)

\(\Leftrightarrow2A=2+2^3+2^4+2^5+...+2^{2019}\)

\(\Leftrightarrow2A-A=\left(2+2^{2019}\right)-\left(1+2^2\right)\)

\(\Leftrightarrow A=2+2^{2019}-1-2^2\)

\(\Leftrightarrow A=2+2^{2019}-5\)

\(\Leftrightarrow A=2^{2019}-3\)

Vậy \(A=2^{2019}-3\).

b) Đặt:

\(B=1+5+5^2+5^3+...+5^{2017}\)

\(\Leftrightarrow5B=5+5^2+5^3+5^4+...+5^{2018}\)

\(\Leftrightarrow5B-B=5^{2018}-1\)

\(\Leftrightarrow4B=5^{2018}-1\)

\(\Leftrightarrow B=\dfrac{5^{2018}-1}{4}\)

Vậy \(B=\dfrac{5^{2018}-1}{4}\).

Chúc bạn học tốt!

a)A= 1 + 2²+2³ + 2⁴ +....+2²⁰¹⁸

2A = 2² + 2³ + 2⁴ +......+2²⁰¹⁸ + 2²⁰¹⁹

_

A= 1 + 2²+2³ + 2⁴ +....+2²⁰¹⁸

A= 2^{2019 _{- 1}}

hbewjfewi

Câu 3 = (5 mũ 51 - 1) : 4