a) \(5+5^2+5^3+....+5^{100}\)

đặt \(A=5+5^2+5^3+....+5^{100}\) ( \(A\) có \(100\) số hạng )

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\) ( có \(100\div2=50\) nhóm )

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

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

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

vì \(6⋮6\Rightarrow6\left(5+5^3+....+5^{99}\right)⋮6\Rightarrow A⋮6\)

b) \(2+2^2+2^3+....+2^{100}\)

đặt \(B=2+2^2+2^3+....+2^{100}\) ( \(B\) có \(100\) số hạng )

\(B=\left(2+2^2+2^3+2^4+2^5\right)+.....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\) ( có \(100\div5=20\) nhóm )

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

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

\(B=31\left(2+...+2^{96}\right)\)

vì \(31⋮31\Rightarrow31\left(2+...+2^{96}\right)\Rightarrow B⋮31\)

a) 5+5^2+5^3..+5^100

=(5+5^2)+(5^3+5^4)+....+(5^99+5^100)

=5.(1+5)+5^3.(1+5)+....+5^99.(1+5)

=5.6+5^3.6+.....+5^99.6

=6.(5+5^3+.....+5^99):6

cau b tuong tu nhe ban

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.

ủng hộ mình lên 360 điểm nha các bạn

a)A=2+2^2+2^3+...+2^60 chia hết cho 15

=>(2+2^2+2^3+2^4)+...+(2^57+2^58+2^59+2^60)

=>2.(1+2+2^2+2^3)+...+2^57+(1+2+2^2+2^3)

=>2.15+...+2^57.15

Vì 15 chia hết choo 15

=>a chia hết cho 15

b)B=1+5+5^2+5^3+...+5^56+5^59+5^98 chia hết cho 31

=>(1+5+5^2)+...+5^56.(1+5+5^2)

=>31+....+5^56.3vi2 31 chia hết cho 31

=>B chia hết cho 31

Ta có : 
=2+2^2+2^3+...+2^60 = 2(1+2+2^2+2^3) + 2^5(1+2+2^2+2^3) + ... + 2^57(1+2+2^2+2^3) 
A=(2+2^5+...+2^57)*15 chia het cho 15 

Ai tick mình đi cho tròn 20 điểm

Sửa đề: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

a: Đặt \(B=2+2^2+2^3+\cdots+2^{10}\)

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

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

\(=31\cdot\left(2+2^6\right)\) ⋮31

Đặt \(C=1+5+5^2+\cdots+5^{14}\)

\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+\cdots+\left(5^{12}+5^{13}+5^{14}\right)\)

\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+\cdots+5^{12}\left(1+5+5^2\right)\)

\(=31\left(1+5^3+\cdots+5^{12}\right)\) ⋮31

Ta có: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

=B+C

mà B⋮31 và C⋮31

nên A⋮31

b: Ta có: \(B=2+2^2+2^3+\cdots+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^9+2^{10}\right)\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+\cdots+2^8\left(2+2^2\right)\)

\(=6\left(1+2^2+\cdots+2^8\right)\) ⋮6

Ta có: \(C=1+5+5^2+\cdots+5^{14}\)

\(=1+\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{13}+5^{14}\right)\)

\(=1+5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{13}\left(1+5\right)\)

\(=1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>C chia 6 dư 1

Ta có: A=B+C

\(=6\left(1+2^2+\cdots+2^8\right)+1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>A chia 6 dư 1

ko biết