Lời giải:
Đặt $A=1-2+2^2-2^3+2^4-2^5+2^6-....-2^{2021}+2^{2022}$

$A=1+(-2+2^2-2^3)+(2^4-2^5+2^6)+(-2^7+2^8-2^9)+...+(2^{2020}-2^{2021}+2^{2022})$

$A=1+(-2+2^2-2^3)+2^3(2-2^2+2^3)+2^6(-2+2^2-2^3)+....+2^{2019}(2-2^2+2^3)$

$=1+(-6)+2^3.6+2^6(-6)+....+2^{2019}.6$

$=1+6(-1+2^3-2^6+...+2^{2019})$

Suy ra $A$ chia $6$ dư $1$/

\(A=5+4^2+...+4^{2021}\\ A=4^0+4^1+...+4^{2021}\\ 4A=4^1+4^2+...+4^{2022}\\ 4A-A=\left(4^1+4^2+...+4^{2022}\right)-\left(4^0+4^1+...+4^{2021}\right)\\ 3A=4^{2022}-1\\ 3A+1=4^{2022}⋮4^{2021}\)

Lời giải:
$A-1=4+4^2+4^3+...+4^{2020}+4^{2021}$
$4(A-1)=4^2+4^3+4^4+....+4^{2021}+4^{2022}$

$\Rightarrow 4(A-1)-(A-1)=4^{2022}-4$

$3(A-1)=4^{2022}-4$

$\Rightarrow 3A+1=4^{2022}\vdots 4^{2021}$ 

TL:

A)   \(A=5+5^2+5^3+5^4+...+5^{49}+5^{50}\)

      \(5.A=5\left(5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)

       \(5A=5^2+5^3+5^4+...+5^{50}+5^{51}\)

        \(5A-A=\left(5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(5+5^2+5^3+5^4+...+5^{49}+5^{50}\right)\)

         \(4A=5^{51}-5\)

Vậy \(4A=5^{51}-5\left(đpcm\right)\)

B)      \(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{49}+5^{50}\right)\)

          \(A=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{49}\left(1+5\right)\)

          \(A=5.6+5^3.6+...+5^{49}.6\)

           \(A=6.\left(5+5^3+...+5^{49}\right)⋮6\)

Vậy \(A\)chia hết cho 6 

HT!!~!

a: =(-1)+(-1)+...+(-1)=-1011

b: =(-5)+(-5)+...+(-5)=-175

a: M=-2021+2021-68-68+17

=-119

b: B=(-1)+(-1)+...+(-1)

=-1x500

=-500

c: C=(1-2-3+4)+(5-6-7+8)+...+(997-998-999+1000)

=0