Ta có \(B=2^1+2^2+2^3+...+2^{30}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{31}\)

\(\Rightarrow B=2B-B=\)\(\left(2^2+2^3+2^4+...+2^{31}\right)-\left(2^1+2^2+2^3+...+2^{30}\right)\)

\(\Leftrightarrow B=2^{31}-2=2\left(2^{30}-1\right)=2\left(8^{10}-1\right)\)

Mà \(8^{10}-1⋮\left(8-1\right)\Leftrightarrow8^{10}-1⋮7\) (1)

Mặt khác \(8^{10}-1=\left(9-1\right)^{10}-1=BS3+1-1=BS3\left(2\right)\)

(1) ; (2) và (7;3) = 1 \(\Rightarrowđpcm\)

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B=(2+2^2+2^3+2^4+2^5+2^6)+...(2^30+2^29+2^28+2^27+2^26+2^25)

B=(2+2^2+2^3+2^4+2^5+2^6)+...+2^25(2+2^2+2^3+2^4+2^5+2^6)

B= 126+...+2^25*126

B=21*6(1+...+2^25) chia hết cho 21

suy ra b chia hết cho 21

kl: vậy B=.... chia hết cho 21

Ta có:

B=1+2+2²+2³+...+2¹⁴⁹⁹

B=(1+2)+(2²+2³)+...+(2¹⁴⁹⁸+2¹⁴⁹⁹)

B=3+2²(1+2)+...+2¹⁴⁹⁸(1+2)

B=3+2².3+...+2¹⁴⁹⁸.3

B=3(1+2²+...+2¹⁴⁹⁸)

\(\Rightarrow B⋮3\)

Vậy\(B⋮3\)

B = 3¹ + 3² + 3³ + ... + 3²⁸ + 3²⁹ + 3³⁰

B = (  3¹ + 3² + 3³ ) + ... + ( 3²⁸ + 3²⁹ + 3³⁰ )

B = 3¹ . ( 1 + 3 + 3² ) + ... + 3²⁸ . ( 1 + 3 + 3² )

B = 3¹ . 13 + ... + 3²⁸ . 13

B = 13 . ( 3 + ... + 3²⁸ ) \(⋮\)13

Vậy B \(⋮\)13 ( dpcm )

\(B=3^1+3^2+3^3+3^4+3^5+............+3^{30}\)

\(\Rightarrow B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+............+\left(3^{28}+3^{29}+3^{30}\right)\)

\(\Rightarrow B=3^1.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+.........+3^{28}.\left(1+3+3^2\right)\)

\(\Rightarrow B=3^1.13+3^4.13+.........+3^{28}.13\)

\(\Rightarrow B=13\left(3^1+3^4+.........+3^{28}\right)\)

Mà 13 \(⋮\)13 \(\Rightarrow13\left(3^1+3^4+...........+3^{28}\right)⋮13\)

Vậy B chia hết cho 13

Ta có : 21 = 3 . 7

Mà 3 và 7 nguyên tố cùng nhau

=> B \(⋮\)21 khi B \(⋮\)3 và B \(⋮\)7

• \(B=2+2^2+2^3+...+2^{30}\)

\(\Rightarrow B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{29}+2^{30}\right)\)

\(\Rightarrow B=2.\left(2^0+2^1\right)+2^3.\left(2^0+2^1\right)+...+2^{29}.\left(2^0+2^1\right)\)

\(\Rightarrow B=2.3+2^3.3+...+2^{29}.3\)

\(\Rightarrow B=3.\left(2+2^3+...+2^{29}\right)⋮3\)( 1 )

• \(B=2+2^2+2^3+...+2^{30}\)

\(\Rightarrow B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{28}+2^{29}+2^{30}\right)\)

\(\Rightarrow B=2.\left(2^0+2^1+2^2\right)+2^4.\left(2^0+2^1+2^2\right)+...+2^{28}.\left(2^0+2^1+2^2\right)\)

\(\Rightarrow B=2.7+2^4.7+...+2^{28}.7\)

\(\Rightarrow B=7.\left(2+2^4+...+2^{28}\right)⋮7\)( 2 )

Từ ( 1 ) và ( 2 ) 

\(\Rightarrow B⋮3.7\)

\(\Rightarrow B⋮21\)

B= 2+ 2²+ 2³+...+ 2²⁹+ 2³⁰.

B có số các số hạng là:

( 30- 1): 1+ 1= 30( số)

* Ta ghép 2 số hạng vào 1 nhóm được 15 nhóm.

=> B=( 2+ 2²)+( 2³+ 2⁴)+( 2⁵+ 2⁶)+...+( 2²⁷+ 2²⁸)+( 2²⁹+ 2³⁰).

=> B= 2( 1+ 2)+ 2³( 1+ 2)+ 2⁵( 1+ 2)+...+ 2²⁷( 1+ 2)+ 2²⁹( 1+ 2).

B= 2. 3+ 2³. 3+ 2⁵. 3+...+ 2²⁷. 3+ 2²⁹. 3.

B= 3( 2+ 2³+ 2⁵+...+ 2²⁷+ 2²⁹)\(⋮\)3.

=> B\(⋮\) 3( 1)

* Ta ghép 3 số hạng vào 1 nhóm được 10 nhóm.

=> B=( 2+ 2²+ 2³)+( 2⁴+ 2⁵+ 2⁶)+( 2⁷+ 2⁸+ 2⁹)+...+( 2²⁵+ 2²⁶+ 2²⁷)+( 2²⁸+ 2²⁹+ 2³⁰).

B= 2( 1+ 2+ 2²)+ 2⁴( 1+ 2+ 2²)+ 2⁷( 1+ 2+ 2²)+...+ 2²⁵( 1+ 2+ 2²)+ 2²⁸( 1+ 2+ 2²).

B= 2. 7+ 2⁴. 7+ 2⁷. 7+...+ 2²⁵. 7+ 2²⁸. 7.

B= 7( 2+ 2⁴+ 2⁷+...+ 2²⁵+ 2²⁸)\(⋮\) 7.

=> B\(⋮\) 7( 2).

( 3; 7)= 1( 3).

Từ( 1);( 2);( 3)

=> B\(⋮\) 21.

=> đpcm.

a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A< 2-\frac{1}{50}\)

\(A< 2\)

b, \(B=2+2^2+2^3+...+2^{30}\)

Ta có :\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{29}+2^{30}\right)\)

\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{29}\left(1+2\right)\)

\(B=2.3+2^3.3+...+2^{29}.3\)

\(B=3\left(2+2^3+...+2^{29}\right)\)chia hết cho 3(1)

Lại có\(B=\left(2+2^2+2^4\right)+...+\left(2^{28}+2^{29}+2^{30}\right)\)

\(B=2\left(1+2+4\right)+...+2^{28}\left(1+2+4\right)\)

\(B=2.7+...+2^{28}.7\)

\(B=7\left(2+...+2^{29}\right)\) chia hết cho 7 (2) 

Mà (3,7)=1 (3) 

Từ (1)(2)(3) => B chia hết cho 21