a) A=2¹+2²+2³+...+2²⁰¹⁰

    A=(2¹+2²)+(2³+2⁴)+.....+(2²⁰⁰⁹+2²⁰¹⁰)

    A=(2¹x3)+(2³x3)+.....+(2²⁰⁰⁹x3)

    A=3x(2¹+2³+.......+2²⁰⁰⁹)

Vậy A chia hết cho 3.

NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !

A= 3( 1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8) chia hết cho 3

mk ko hieu ban co the giai thich ra duoc ko

giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!

Câu b, bài b1 chứng minh \(a=2^{2006}-1?\)

a: \(=\dfrac{4}{9}\left(4+\dfrac{1}{3}+13+\dfrac{2}{3}\right)=\dfrac{4}{9}\cdot18=8\)

b: \(=\dfrac{3}{8}\left(5+\dfrac{1}{4}+10+\dfrac{3}{4}\right)=\dfrac{3}{8}\cdot16=6\)

c: \(=\dfrac{-2}{7}\left(6+\dfrac{1}{5}+14+\dfrac{4}{5}\right)=\dfrac{-2}{7}\cdot21=-6\)

a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=\dfrac{35}{8}+\dfrac{17}{3}\)

\(=\dfrac{105}{24}+\dfrac{136}{24}\)

\(=\dfrac{241}{24}\)

b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)

\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)

\(=\dfrac{29}{8}+\dfrac{27}{7}\)

\(=\dfrac{419}{56}\)

c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)

\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)

\(=\dfrac{9}{8}+\dfrac{16}{3}\)

\(=\dfrac{155}{24}\)

d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)

\(=\dfrac{17}{6}:\dfrac{1}{2}\)

\(=\dfrac{17}{6}\cdot2\)

\(=\dfrac{17}{3}\)

e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)

\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)

\(=\dfrac{39}{4}-\dfrac{6}{7}\)

\(=\dfrac{249}{28}\)

a: =4+3/8+5+2/3

=9+9/24+16/24

=9+25/24

=216/24+25/24=241/24

b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)

=27/7+29/8

=419/56

c: =2+3/8-1-1/4+5+1/3

=6+3/8-1/4+1/3

=6+3/8+1/12

=144/24+9/24+2/24

=155/24

d: =(15/6+2/6):1/2

=17/6*2

=17/3

e: =(15/6-2/6)*9/2-6/7

=13/6*9/2-6/7

=117/12-6/7

=249/28

\(4A=4+4^2+4^3+4^4+4^5+4^6+4^7\\ 4A-A=4^7-1\\ 3A=4^7-1\\ A=\dfrac{4^7-1}{3}44\)

ta có:\(A=1+4+...+4^6\Rightarrow4A=4+4^2+...+4^7\)

\(\)\(4A=4+4^2+..+4^7\\ \Rightarrow4A-A=\left(4+4^2+..+4^7\right)-\left(1+4+...+4^6\right)\\ \Rightarrow3A=4^7-1\\ \Rightarrow A=\dfrac{4^7-1}{3}\)

\(C=1+4+4^2+...+4^6\)

\(4C=4+4^2+...+4^7\)

\(4C-C=\left(4+4^2+...+4^7\right)-\left(1+4+4^2+...+4^6\right)\)

\(3C=4^7-1\)

\(\Rightarrow C=\dfrac{4^7-1}{3}\left(đpcm\right)\)