Bài 3: 

a) Ta có: \(C=2+2^2+2^3+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\cdot\left(2+2^6+...+2^{96}\right)⋮31\)(đpcm)

Bài 1: 

Ta có: \(A=3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n\cdot9-2^n\cdot4+3^n-2^n\)

\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

Vậy: A có chữ số tận cùng là 0

Bài 2: 

Ta có: \(abcd=1000\cdot a+100\cdot b+10\cdot c+d\)

\(\Leftrightarrow abcd=1000\cdot a+96\cdot b+8c+2c+4b+d\)

\(\Leftrightarrow abcd=8\left(125a+12b+c\right)+\left(2c+4b+d\right)\)

mà \(8\left(125a+12b+c\right)⋮8\)

và \(2c+4b+d⋮8\)

nên \(abcd⋮8\)(đpcm)

a) Tổng A có số số hạng là:

`(101-1):1+1=101`(số hạng)

b) `A=2+2^3 +2^5 +...+2^101`

`2^2 A=2^3 +2^5 +2^7 +...+2^103`

`4A-A=2^3 +2^5 +2^7 +...+2^103 -2-2^3 -2^5 -...-2^101`

`3A=2^103 -2`

`=>3A+2=2^103 -2+2=2^103`

c) `A=2+2^3 +2^5 +...+2^101`

`A=2(1+2^2 +2^4 +...+2^100)⋮2`

`A=2+2^3 +2^5 +...+2^101`

`A=2(1+2^2 +2^4)+...+2^97 .(1+2^2 +2^4)`

`A=2.21+...+2^97 .21`

`A=21(2+...+2^97)⋮21`

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

Bài 1

a, cm : A = 16⁵ + 2¹⁵ ⋮ 3

    A = 16⁵ + 2¹⁵

   A = (2⁴)⁵ +  2¹⁵

  A  = 2²⁰ + 2¹⁵

 A  =  2¹⁵.(2⁵ + 1)

 A = 2¹⁵. 33 ⋮ 3 (đpcm)

b,cm : B = 8⁸ + 2²⁰ ⋮ 17

    B = (2³)⁸ + 2²⁰ 

    B =  2¹⁶ + 2²⁰

    B = 2¹⁶.(1 + 2⁴)

    B = 2¹⁶. 17 ⋮ 17 (đpcm)

c, cm: C = 1 - 2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶ -...-2²⁰²¹ + 2²⁰²² : 6 dư 1

C=1+(-2+2²-2³+2⁴- 2⁵+2⁶)+...+(-2²⁰¹⁷+2²⁰¹⁸-2²⁰¹⁹+2²⁰²⁰-2²⁰²¹+2²⁰²²)

C = 1 + 42 +...+ 2²⁰¹⁶.(-2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶)

C = 1 + 42+...+ 2²⁰¹⁶.42

C = 1 + 42.(2⁰+...+2²⁰¹⁶)

42 ⋮ 6 ⇒ C = 1 + 42.(2⁰+...+2²⁰¹⁶) : 6 dư 1 đpcm

a, \(\overline{aaa}\) \(⋮\) 37

    \(\overline{aaa}\) = a x 111 = a x 3 x 37 ⋮ 37 (đpcm)

b, (\(\overline{ab}\) + \(\overline{ba}\)) ⋮ 11

  \(\overline{ab}\) + \(\overline{ba}\) = \(\overline{a0}\) + b + \(\overline{b0}\) + a = \(\overline{aa}\) + \(\overline{bb}\) = a x 11 + b x 11 = 11 x (a+b)⋮11

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

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 $0$