Lời giải:

$B=3+3^2+(3^3+3^4+3^5)+(3^6+3^7+3^8)+....+(3^{1989}+3^{1990}+3^{1991})$

$=12+3^3(1+3+3^2)+3^6(1+3+3^2)+...+3^{1989}(1+3+3^2)$

$=12+(1+3+3^2)(3^3+3^6+...+3^{1989})$

$=12+13(3^3+3^6+...+3^{1989})$

$\Rightarrow B$ chia $13$ dư $12$.

2/

$B=3+3^2+3^3+...+3^{1991}$

$3B=3^2+3^3+3^4+...+3^{1992}$
$\Rightarrow 3B-B=3^{1992}-3$

$\Rightarrow 2B=3^{1992}-3$

Có:

$3^4\equiv -1\pmod {41}$

$\Rightarrow 3^{1992}=(3^4)^{498}\equiv (-1)^{498}\equiv 1\pmod {41}$

$\Rightarrow 3^{1992}-3\equiv 1-3\equiv -2\pmod {41}$

$\Rightarrow 2B\equiv -2\pmod {41}$

$\Rightarrow 2B\not\vdots 41$

$\Rightarrow B\not\vdots 41$.

a)\(B=3+3^2+3^3+....+3^{30}\)

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{31}\)

\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{31}\right)-\left(3+3^2+3^3+....+3^{30}\right)\)

\(\Rightarrow2B=3^{31}-3\)

\(\Rightarrow B=\frac{3^{31}-3}{2}\)

b) \(B=3+3^2+3^3+3^4+...+3^{30}\)

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

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

             \(=4.\left(3+3^3+.....+3^{29}\right)⋮4\)

Vậy B chia hết cho 4

                                                     Bài giải

              Ta có : 

\(1+3+3^2+3^3+3^4+...+3^9\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{98}+3^{99}\right)\)

\(=4+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^{98}\left(1+3\right)\)

\(=4+3^2\cdot4+3^4\cdot4+...+3^{98}\cdot4\)\(⋮\text{ }4\)

\(\Rightarrow\text{ ĐPCM}\)

                                       Bài giải

      \(1+3+3^2+3^3+3^4+...+3^9\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{98}+3^{99}\right)\)

\(=4+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^{98}\left(1+3\right)\)

\(=4+3^2\cdot4+3^4\cdot4+...+3^{98}\cdot4\)\(⋮\text{ }4\)

\(\Rightarrow\text{ ĐPCM}\)

      \(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)

\(=\frac{7}{4}\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\right)\)

\(=\frac{7}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}\)

\(=\frac{35}{21}\)

b: B=3(1+3)+3^3(1+3)+...+3^2009(1+3)

=4(3+3^3+...+3^2009) chia hết cho 4

B=3(1+3+3^2)+3^4(1+3+3^2)+...+3^2008(1+3+3^2)

=13(3+3^4+...+3^2008) chia hết cho 13

c: \(C=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2009}\left(1+5\right)\)

\(=6\left(5+5^3+...+5^{2009}\right)⋮6\)

\(C=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)

\(=31\left(5+5^4+...+5^{2008}\right)⋮31\)

d: \(D=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{2009}\left(1+7\right)\)

\(=8\left(7+7^3+...+7^{2009}\right)⋮8\)

\(D=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)

\(=57\left(7+7^4+...+7^{2008}\right)⋮57\)

ử dụng phương pháp phản chứng 
giả sử n chia hết cho 5 
=>n có dạng 5k 
=>n^2+n+1=25k^2+5k+1=5k(5k+1)+1 
ta có 5k(5k+1) chia hết cho 5 mà 1 ko chia hết cho 5 
=>25k^2+5k+1 ko chia hết cho 5 (đpcm)

a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

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

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

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

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

d) Đặt \(D=75+\left(4^{2006}+4^{2005}+4^{2004}+...+1\right).25\)

Đặt \(E=4^{2006}+4^{2005}+4^{2004}+...+1\)

\(\Rightarrow4E=4^{2007}+4^{2006}+4^{2005}+...+4\)

\(\Rightarrow3E=4E-E\)

\(=\left(4^{2007}+4^{2006}+4^{2005}+...+4\right)-\left(4^{2006}+4^{2005}+4^{2004}+...+1\right)\)

\(=4^{2007}-1\)

\(\Rightarrow E=\dfrac{\left(4^{2007}-1\right)}{3}\)

\(\Rightarrow D=75+\dfrac{4^{2007}-1}{3}.25\)

Ta có:

\(4^{2007}=\left(4^2\right)^{1003}.4\)

\(4^2\equiv6\left(mod10\right)\)

\(\left(4^2\right)^{1003}\equiv6^{1003}\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow4^{2007}\equiv\left(4^2\right)^{1003}.4\left(mod10\right)\equiv6.4\left(mod10\right)\equiv4\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(4^{2007}\) là 4

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 = 1 + 3 + 3² +......+ 3¹¹

   = (1+3)+(3²+3³)+.....+(3¹⁰+3¹¹)

   = 1.(1+3)+3²(1+3)+.....+3¹⁰(1+3)

   = (1+3)(1+3²+.....+3¹⁰)

   = 4(1+3²+......+3¹⁰) chia hết cho 4

Vậy B chia hết cho 4

câu b của bạn thiếu số 3 ở giữa số 1 và 3² nghen