\(S=2+2^2+2^3+2^4+....+2^{99}+2^{100}\)

\(S=2.\left(2+2^2\right)+.....+2^{99}.\left(2+2^2\right)\)

\(S=2.6+.....+2^{99}.6\)

\(S=6.\left(2+2^{99}\right)⋮6\)

\(\Rightarrow S⋮6\)

ta có :

2 + 2² + 2³ + ....... + 2⁹⁹ + 2¹⁰⁰ = a

2¹ + 2² + ...+ 2¹⁰⁰  + 2¹⁰¹ = 2a

=> a = 2¹⁰¹ -   2 

( hình như vậy , dạng rút gọn này mik chỉ nhớ máng máng , sai thôi xin thứ lỗi )

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) \(S=1+4+4^2+4^3+...4^{62}\)

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

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

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

\(\Rightarrow dpcm\)

\(S=2^0+2^2+...+2^{2014}.\)

\(S=\left(2^0+2^2+2^4+2^6\right)+.....+\left(2^{2008}+2^{2010}+2^{2012}+2^{2014}\right)\)

\(S=17+.....+2^{2008}.17\)

\(S=17.\left(2^0+...+2^{2008}\right)\)

\(\Leftrightarrow S⋮17\left(đpcm\right)\)

\(S=2^0+2^2+...+2^{2014}.\)

\(S=\left(2^0+2^2+2^4\right)+....+\left(2^{2010}+2^{2012}+2^{2014}\right)\)

\(S=21+....+2^{2010}.21\)

\(S=21.\left(2^0+...+2^{2010}\right)\)

\(S=7.3.\left(2^0+....+2^{2010}\right)\)

\(\Leftrightarrow S⋮7\left(đpcm\right)\)

S = 2⁰ + 2² + 2⁴ + 2⁶ + 2⁸ + ... + 2²⁰¹⁴

S = (2⁰ + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁰ + 2²⁰¹² + 2²⁰¹⁴)

S = (2⁰ + 2² + 2⁴) + 2⁶(2⁰ + 2² + 2⁴) + ... + 2²⁰¹⁰(2⁰ + 2² + 2⁴)

S = (2⁰ + 2² + 2⁴)(2⁶ + ... + 2²⁰¹⁰) 

S =         21   .  (2⁶ + ... + 2²⁰¹⁰) 

Vì 21 \(⋮\)7 nên 21 . (2⁶ + ... + 2²⁰¹⁰)  \(⋮\)7 => S \(⋮7\)

S=2⁰+2²+...+2²⁰¹⁴.

S=(2⁰+2²+2⁴+2⁶)+.....+(2²⁰⁰⁸+2²⁰¹⁰+2²⁰¹²+2²⁰¹⁴)

S=17+.....+2²⁰⁰⁸.17

S=17.(2⁰+...+2²⁰⁰⁸)

=> S \(⋮\)17

\(S=3^0+3^2+3^4+3^6+...+3^{2014}\)

\(=1+3^2+3^4+3^6+...+3^{2014}\)

\(=\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^{2012}\left(1+3^2\right)\)

\(=7+3^4.7+...+3^{2012}.7=7\left(1+3^4+...+3^{2012}\right)⋮7\)

Vậy ta có đpcm 

b) S = 3⁰ + 3² + 3⁴ + .. + 3²⁰¹⁴

S = (3⁰ + 3² + 3⁴) + (3⁶ + 3⁸ + 3¹⁰) + ... + (3²⁰¹⁰ + 3²⁰¹² + 3²⁰¹⁴)

S = 3⁰(1 + 3² + 3⁴) + 3⁶.(1 + 3² + 3⁴) + ... + 3²⁰¹⁰.(1 + 3² + 3⁴)

S = 3⁰.91 + 3⁶.91 + ... + 3²⁰¹⁰.91

S = 91.(3⁰ + 3⁶ + .. + 3²⁰¹⁰) = 7.13.(3⁰ + 3⁶ + .. + 3²⁰¹⁰)

Vì tích trên có thừa số 7 => S chia hết cho 7 

S = 3⁰ + 3² + 3⁴ + ... + 3²⁰¹⁴

=> 3². S = 3².(3⁰ + 3² + 3⁴ + ... + 3²⁰¹⁴) = 3² + 3⁴ + 3⁶ + ... + 3²⁰¹⁶

=> 3².S - S = (3² + 3⁴ + 3⁶ + ... + 3²⁰¹⁶) - (3⁰ + 3² + 3⁴ + ... + 3²⁰¹⁴) = 3²⁰¹⁶ - 1 

=> 8.S = 3²⁰¹⁶ - 1 

=> S = \(\frac{3^{2016}-1}{8}\)

 a ) S = 2⁰ +2² + 2⁴ +...+ 2²⁰¹⁴     

    4S = 2² + 2⁴ + 2⁶ + ... + 2²⁰¹⁶

Mà S =  ( 4S- S ) : 3 

=>  S = [ ( 2² + 2⁴  + 2⁶ +...+ 2²⁰¹⁶ ) - ( 2⁰ + 2² + 2⁴ +...+ 2²⁰¹⁴ ) ] : 3

          = [ 2²⁰¹⁶ - 2⁰  ]   : 3

          = \(\frac{2^{2016}-1}{3}\)    

b) S = 2⁰ + 2² + 2⁴ + ... + 2²⁰¹⁴

       = ( 2⁰ + 2² + 2⁴ ) + ( 2⁵ + 2⁶ + 2⁷ ) + ...+ ( 2²⁰¹⁰ + 2²⁰¹² + 2²⁰¹⁴ )

       =    21   +  2⁵ x ( 2⁰ + 2² + 2⁴ ) +... + 2²⁰¹⁰ x  ( 2⁰ + 2² + 2⁴ )

       =   21 +  2⁵ x 21   + ... + 2²⁰¹⁰ x 21

       = 21 x  ( 1 + 2⁵ + ... + 2²⁰¹⁰ )

=> S \(⋮\)21    (đpcm)

a) S= 2 + 2² + 2³ +...+ 2¹⁰⁰

S= ( 2+2² ) + ( 2³+2⁴ ) +...+( 2⁹⁹ + 2¹⁰⁰ )

S= 6+ 2² ( 2+2²)+ ...+ 2⁹⁸ (2+2²)

S=6+ 2².6+ ...+ 2⁹⁸.6

S= 6.(2²+...+2⁹⁸) chia hết cho 3 ( vì 6 chia hết cho 3)