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

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

S = 2.3+2 ^3 .3+...+2 ^99 .3

S = 3.(2+2^ 3+...+2^ 99 ) =>

S chia hết cho 3

S = (2^ 1+2^ 2+2^ 3+2 ^4 )+(2^ 5+2^ 6+2^ 7+2 ^8 )+...+(2^ 97+2^ 98+2^ 99+2 ^100 )

S = 2.(1+2+4+16)+2^ 5 .(1+2+4+16)+...+2^ 97 .(1+2+4+16) S = 2.15+2^ 5 .15+...+2^ 97 .15

S = 15.(2+2^ 5+...+2^ 97 ) =>

S chia hết cho 15 

\(H=2+2^2+...+2^{100}\)

\(\Rightarrow H=\left(2+2^2+2^3+2^4\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(\Rightarrow H=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(\Rightarrow H=2.15+...+2^{97}.15\)

\(\Rightarrow H=\left(2+...+2^{97}\right).15⋮15\)

\(\Rightarrow H⋮15\left(đpcm\right)\)

minh chi lam dc cau a thoi nha nhung hay t i c k cho minh

3 + 3² = 12 chia het cho 4  3 + 3² + 3³ + .......+3⁹ + 3¹⁰ = 3⁰ .[ 3+3² ] + 3² . [ 3 + 3² ] + ....+3⁸ . [ 3 + 3² ]

=3⁰ . 12 + 3²  . 12 +.....+ 3⁸ . 12 = 12.[3⁰ + 3² +....+ 3⁸ ] 

vi 12 chia het cho 4 nen 12 nhan voi so tu nhien nao thi so do cung chia het cho 4 nen A chia het cho 4

hghjhgjhgjh

a) \(\left(3+3^2\right)+...+\left(3^9+3^{10}\right)\)

 \(\Rightarrow3.\left(1+3\right)+...+3^9.\left(1+3\right)\)

\(\Rightarrow4.\left(3+...+3^9\right)⋮4\)

b)\(H⋮155\Leftrightarrow H⋮31;5\)

gộp 4 số  chia hết cho 5 (1)

gộp 5 số chia hết cho 31(2)

từ (1)và(2) suy ra H chia hết cho 155