A=5+5²+...+5⁹⁹+5¹⁰⁰

=(5+5²)+...+(5⁹⁹+5¹⁰⁰)

=5.(1+5)+...+5⁹⁹.(1+5)

=5.6+...+5⁹⁹.6

=6.(5+...+5⁹⁹) chia hết cho 6 (dpcm)

Ccá câu khcs bạn cứ dựa vào câu a mà làm vì cách làm tương tự chỉ hơi khác 1 chút thôi

Chúc bạn học giỏi nha!!

\(A=5+5^2+5^3+...+5^{100}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{99}+5^{100}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)

\(=5.6+5^3.6+...+5^{99}.6\)

\(=6\left(5+5^3+...+5^{99}\right)⋮6\)(đpcm)

\(B=2+2^2+2^3+...+2^{100}\)

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

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

\(=2.31+...+2^{96}.31\)

\(=31\left(2+...+9^{96}\right)⋮31\)(đpcm)

\(C=3+3^2+3^3+...+3^{60}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)

\(=3.4+3^3.4+...+3^{59}.4\)

\(=4\left(3+3^3+...+3^{59}\right)⋮4\)(đpcm)

\(C=3+3^2+3^3+...+3^{60}\)

\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(=3.13+...+3^{58}.13\)

\(=13\left(3+...+3^{58}\right)⋮13\)(đpcm)

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)

B = 3¹ + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰

3B = 3(3¹ + 3² + 3³ + ..... + 3⁹⁹ + 3¹⁰⁰)

3B = 3² + 3³ + 3⁴ +...... + 3¹⁰⁰ + 3¹⁰¹

3B - B = 2B = (3² + 3³ + 3⁴ + .... + 3¹⁰⁰ + 3¹⁰¹) - ( 3¹ + 3² + 3³ + .... + 3¹⁰⁰)

2B = (3² - 3²) + (3³ - 3³) +.....+ ( 3¹⁰⁰ - 3¹⁰⁰) + ( 3¹⁰¹ - 1)

2B = 0 + 0 + 0 + ..... +0 + 3¹⁰¹ - 1

2B = 3¹⁰¹ - 1

B = (3¹⁰¹ - 1)  : 2

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

A=2^1+2^2+2^3+2^4+...+2^2010 

=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)

=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)

=2.3+2^3.3+...+2^2010.3

=(2+2^3+2^2010).3

=> A chia het cho 3

​​​​ 

Mà câu c bạn đánh chia hết thành chết hết rồi kìa

em chịu!!!!!!!!!!!

B = 3^x+3 + 3^x+1 + 2^x+3 + 2^x+2

   = 3^x.3³ + 3^x.3 + 2^x.2³ + 2^x.2²

   = 3^x(3³ + 3) + 2^x(2³ + 2²)

   = 3^x.30 + 2^x.12

Vì 3^x.30 chia hết cho 6     => 3^x.30 + 2^x.12 chia hết cho 6

    2^x.12 chia hết cho 6

=> B chia hết cho 6

\(C=\left(2+2^2+...+2^4\right)+\left(2^5+...+2^8\right)+...+\left(2^{97}+...+2^{100}\right)\text{ chia hết cho 31 (dễ)}\)

\(b,2C=4+2^3+....+2^{101}\text{ do đó: }2C-C=C=2^{101}-2=2^{2x-1}-2\text{ do đó:}x=101\)

Nhóm thiếu kìa Khải :v 

a) C = 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰

= ( 2 + 2² + 2³ + 2⁴ + 2⁵ ) + ( 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰ ) + ... + ( 2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹ + 2¹⁰⁰ )

= 2( 1 + 2 + 2² + 2³ + 2⁴ ) + 2⁶( 1 + 2 + 2² + 2³ + 2⁴ ) + ... + 2⁹⁶( 1 + 2 + 2² + 2³ + 2⁴ )

= 2.31 + 2⁶.31 + ... + 2⁹⁶.31

= 31( 2 + 2⁶ + ... + 2⁹⁶ ) chia hết cho 31 ( đpcm )

b) C = 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰

2C = 2( 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰ )

= 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹

C = 2C - C 

= 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹ - ( 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰ )

= 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹ - 2 - 2² - 2³ - 2⁴ - 2⁵ - 2⁶ + ... - 2⁹⁹ - 2¹⁰⁰

= 2¹⁰¹ - 2

2^2x-1 - 2 = C

<=> 2^2x-1 - 2 = 2¹⁰¹ - 2

<=> 2^2x-1 = 2¹⁰¹

<=> 2x - 1 = 101

<=> 2x = 102

<=> x = 51

ta có :

`1^3` \(⋮\) `1`

\(2^3⋮2\)

\(3^3⋮3\)

.................

\(100^3⋮100\)

`=>` \(1^3+2^3+3^3+...+100^3⋮1+2+3+...+100\)

vậy `A` \(⋮\)`B`