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 

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

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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

câu 1 :1/100

câu 2 :1

b)ta đặt A:  \(A=\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\)

                   \(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+..+\left(\frac{98}{2}+1\right)+\left(\frac{99}{1}-98\right)\)

                  \(A=\frac{100}{99}+\frac{100}{98}+..+\frac{100}{2}+\frac{100}{100}\)

                  \(A=100\cdot\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}\right)\)

Đặt \(A=2^{100}-2^{99}-2^{98}-2^{97}-\cdot\cdot\cdot-2-1\)

\(=-\left(1+2+\cdot\cdot\cdot+2^{99}+2^{100}\right)\)

Đặt \(B=1+2+\cdot\cdot\cdot+2^{99}+2^{100}\)

\(2B=2+2^2+\cdot\cdot\cdot+2^{100}+2^{101}\)

\(2B-B=2+2^2+\cdot\cdot\cdot+2^{100}+2^{101}-\left(1+2+\cdot\cdot\cdot+2^{99}+2^{100}\right)\)

\(B=2^{101}-1\)

Thay \(B=2^{101}-1\) vào \(A\), ta được:

\(A=-\left(2^{101}-1\right)\)

\(=1-2^{101}\)

#\(Toru\)