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)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

a/A= \(5^6-10^4=5^4.\left(5^2-2^4\right)=5^4.\left(25-16\right)=5^4.9\)chia hết cho 9

b/\(F=5+5^2+5^3+5^4+5^5+5^6=\left(5+5^2+5^3\right).\left(5^4+5^5+5^6\right)=\left(5+25+125\right)\left(5^4+5^5+5^6\right)=155.\left(5^4+5^5+5^6\right)\)

vì 155 chia hết cho 31 đa thức F chia hết cho 31

c)D=4+4²+4³+4⁴+...+4²⁰¹²

D=(4+4²)+(4³+4⁴)+...+(4²⁰¹¹+4²⁰¹²)

D=4.5+4³.5+4⁵.5+...+4²⁰¹¹.5

D=5.(4+4³+4²⁰¹¹)

=>D chia hết cho 5

=>ĐPCM

tất cả đều có trong câu hỏi tương tự

b)

A=(1+5+5²)+(5³+5⁴+5⁵)+...(5⁴⁰²+5⁴⁰³+5⁴⁰⁴)

A=31.1+31.5³+...+31.5⁴⁰²

A=31.(1+5³+...+5⁴⁰²)

=>A chia hết cho 31

=>Đâu phải con ma

b) \(A=1+5+5^1+5^2+5^3+...+5^{71}\)

\(\Rightarrow A=\left(1+5^1+5^2\right)+5^3\left(1+5^1+5^2\right)+...+5^{69}\left(1+5^1+5^2\right)\)

\(\Rightarrow A=31+5^3.31+...+5^{69}.31\)

\(\Rightarrow A=31\left(1+5^3+...+5^{69}\right)⋮31\left(dpcm\right)\)

a) \(A=1+5^1+5^2+5^3+...+5^{71}\)

\(\Rightarrow A=\dfrac{5^{71+1}-1}{5-1}=\dfrac{5^{72}-1}{4}\)

\(4A+x=5^{72}\)

\(\Rightarrow4.\dfrac{5^{72}-1}{4}+x=5^{72}\)

\(\Rightarrow5^{72}-1+x=5^{72}\)

\(\Rightarrow x=1\)

Lời giải:

$n$ không chia hết cho $3$ nên $n=3k+1$ hoặc $n=3k+2$ với $k$ tự nhiên.

Nếu $n=3k+1$:
$A=5^{2n}+5^n+1=5^{2(3k+1)}+5^{3k+1}+1$

$=5^{6k}.25+5.5^{3k}+1$

Vì $5^3\equiv 1\pmod {31}$

$\Rightarrow A\equiv 1^{2k}.25+5.1^k+1\equiv 31\equiv 0\pmod {31}$

$\Rightarrow A\vdots 31$

Nếu $n=3k+2$ thì:

$A=5^{2(3k+2)}+5^{3k+2}+1$

$=5^{6k}.5^4+5^{3k}.5^2+1$

$\equiv 1^{2k}.1.5+1^k.5^2+1\equiv 5+5^2+1\equiv 31\equiv 0\pmod {31}$

$\Rightarrow A\vdots 31$

Từ 2 TH suy ra $A\vdots 31$ (đpcm)