\(S=1+2+2^2+2^3+...+2^{59}\)

\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)

\(S=3+2^2\cdot3+...+2^{58}\cdot3\)

\(S=3\cdot\left(1+2^2+...+2^{58}\right)\)

S chia hết cho 3

_____

\(S=1+2+2^2+...+2^{59}\)

\(S=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{57}+2^{58}+2^{59}\right)\)

\(S=7+7\cdot2^3+...+7\cdot2^{57}\)

\(S=7\cdot\left(1+2^3+...+2^{57}\right)\)

S chia hết cho 7 

_____

\(S=1+2+2^2+2^3+...+2^{59}\)

\(S=\left(1+2+2^2+2^3\right)+\left(2^4+2^5+2^6+2^7\right)+...+\left(2^{56}+2^{57}+2^{58}+2^{59}\right)\)

\(S=15+2^4\cdot15+...+2^{56}\cdot15\)

\(S=15\cdot\left(1+2^4+...+2^{56}\right)\)

S chia hết cho 15 

Đề sai, viết lại thành:

A= 2¹+2²+2³+2⁴+...+2⁵⁹+2⁶⁰

Giải:

A=2¹+2²+2³+...............+2⁵⁹+2⁶⁰

A=(2¹+2²+2³)+...............+(2⁵⁸+2⁵⁹+2⁶⁰)

A=2.(1+2+2²)+............+2⁵⁸.(1+2+2²)

A=2.7+.......................+2⁵⁸.7

A=(2+2⁴+..............+2⁵⁸).7 ⋮ 7(đpcm)

Lời giải:
$S=(2+2^2)+(2^3+2^4)+....+(2^{23}+2^{24})$

$=2(1+2)+2^3(1+2)+....+2^{23}(1+2)$

$=(1+2)(2+2^3+...+2^{23})$

$=3(2+2^3+...+2^{23})\vdots 3$

b.

$S=2+2^2+2^3+...+2^{23}+2^{24}$

$2S=2^2+2^3+2^4+....+2^{24}+2^{25}$

$\Rightarrow 2S-S=2^{25}-2$

$\Rightarrow S=2^{25}-2$

Ta có:

$2^{10}=1024=10k+4$

$\Rightarrow 2^{25}-2=2^5.2^{20}-2=32(10k+4)^2-2=32(100k^2+80k+16)-2$
$=10(320k^2+8k+51)\vdots 10$

$\Rightarrow S$ tận cùng là $0$

\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)

A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259)  chia hết cho 3
=>A  chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7  chia hết cho 7 =>7.(2+...+258)  chia hết cho 7

CHIA HẾT CHO 3 :

A= (2+22)+(23+24)+...+(259+260)

A=2.(1+2)+23.(1+2)+...+259.(1+2)

A=2.3+23.3+...+259.3

A=3.(2+23+...+259)

Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3

=>A chia hết cho 3

dcv

s=[1+2]+[2+2 mũ 2]+...+[2 mũ 6+2 mũ 7]

s=1 nhân [1+2]+2 nhân [1+2]+...+2 mũ 6 nhân [1+2]

s=[1+2] nhân[1+2+...+2 mũ 6

s=3 nhân [1+2+...+2 mũ 6]

=> s chia hết cho 3

a: \(2A=2^2+2^3+...+2^{61}\)

=>A=2^61-2

b: \(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{55}+2^{58}\right)\) chia hết cho 7(1)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)=3\left(2+2^3+...+2^{59}\right)⋮3\left(2\right)\)

Từ (1), (2) suy ra A chia hết cho 21

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

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

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3