S=1+2+2²+2³+...+2²⁰

2S=2+2²+2³+2⁴+...+2²¹  

=>S=2S-S=2²¹-1C

Vậy S=2²¹-1

\(S=1+2+2^2+2^3+...+2^{20}\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{21}\)

\(\Rightarrow2S-S=\left(2+2^2+...+2^{21}\right)-\left(1+2+...+2^{20}\right)\)

\(\Rightarrow S=2^{21}-1\)

S=1+2+2^2+....+2^20

=>2S=2+2^2+2^3+...+2^21

=>2S-S=2^21-1

=>S=2^21-1

\(S=1+2+2^2+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow S=2^{10}-1\)

Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)

Vậy \(S< 5.2^8\)

S=1+2+2^2+2^3+...+2^9​

2S=2+2^2+2^3+...+2^9+2^10

2S-S=(2+2^2+2^3+...+2^9+2^10)-(1+2+2^2+2^3+...+2^9)

S=2^10-1

5.2^8=(2^2+1).2^8=(2^2.2^8)+(1.2^8)=2^10+2^8

Vì 2^10-1<2^10+2^8=> S<5.2^8

Vậy S < 5. 2^8

Ta có: \(S=1+2+2^2+2^3+...+2^9\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{10}\right)-\left(1+2+2^2+...+2^9\right)\)

\(\Rightarrow S=2^{10}-1\)

Mặt khác: \(5.2^8=\left(1+2^2\right).2^8=2^8+2^2.2^8=2^8+2^{10}\)

Vì \(2^{10}-1< 2^8+2^{10}\Rightarrow S< 5.2^8\)

http://123link.pw/j6KCoe

   5⁹⁹ - 4² x 5⁹⁷ - 3² x 5⁹

= 5⁹⁷.(5² - 4²) - 3².5⁹

= 5⁹⁷.(25 - 16) - 3².5⁹

= 5⁹⁷.9 - 9.5⁹

b, 2^y+1 + 2^y+2 = 24

   2^y+1.(1 + 2) = 24

   2^y+1.3 = 24

   2^y+1   = 24: 3

   2^y+1 = 8

   2^y+1 = 2³

  y + 1 = 3

   y = 2 

*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!

dùng  máy tính nha b

chúc hok tốt

mình bắt đầu thấy loạn rồi đấy

\(\dfrac{2^{13}+2^5}{2^{10}+2^5}\)

\(=\dfrac{2^5\cdot\left(2^8+1\right)}{2^5\cdot\left(2^5+1\right)}\)

\(=\dfrac{2^8+1}{2^5+1}\)

\(\dfrac{2^{13}+2^5}{2^{10}+2^5}\)

\(=\dfrac{2^5\left(2^8+1\right)}{2^5\left(2^5+1\right)}\)

\(=\dfrac{2^8+1}{2^5+1}\)\(=\dfrac{257}{33}\)

\(A=2^0+2^1+2^2+2^3+2^4+2^5+\dots+2^{100}\\=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+\dots+(2^{99}+2^{100})+2^0\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+\dots+2^{99}\cdot(1+2)+1\\=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{99}\cdot3+1\\=3\cdot(2+2^3+2^5+\dots+2^{99})+1\)

Vì \(3\cdot(2+2^3+2^5+\dots+2^{99})\vdots3\)

\(\Rightarrow 3\cdot(2+2^3+2^5+\dots+2^{99})+1\) chia \(3\) dư 1

hay số dư của phép chia \(A\) cho \(3\) là \(1\).

A=2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100

A=1 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100

A=1 + (2^1 + 2^2) + (2^3 + 2^4) + ....+(2^99 + 2^100)

A=1 + 2.(1+2) + 2^3.(1+2)+....+2^99.(1+2)

A=1 + 2 . 3 + 2^3 . 3 +....+2^99 . 3

A=1 +3 .(2+2^3+..+2^99)

=> A:3 dư 1