\(A=1+3+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(A=1+3+\left(3^2+3^3+3^4+...+3^{99}+3^{100}\right)\)

\(A=1+3\)

\(A=4\)

→ \(4\) ⋮ 4

⇒ \(A\)⋮\(4\)

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)

Đặt A=\(3^1+3^2+3^3+3^4+...+3^{99}+3^{100}\)

A=\(\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

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

A=\(3^1\cdot4+3^3\cdot4+...+3^{99}\cdot4\)

A=\(4\left(3^1+3^3+...+3^{99}\right)⋮4\left(đpcm\right)\)

Có : 3^1 + 3^2 + 3^3 + 3^4 + ..... + 3^99 + 3^100

= (3^1+3^2) + (3^3+3^4) + .... + (3^99+3^100)

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

= 3^1.4 + 3^3.4 + .... + 3^99.4

= 4.(3^1+3^3+.....+3^99) chia hết cho 4

Đặt \(A=3+3^2+3^3+...+3^{99}+3^{100}\)

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

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

\(\Rightarrow A=4\left(3+3^3+...+3^{99}\right)\)

\(\Rightarrow A⋮4\)

\(\RightarrowĐPCM\)

\(A=3^1+3^2+...+3^{99}+3^{100}\)

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

\(A=3.4+...+3^{99}.4\)

\(A=4.\left(3+...+3^{99}\right)\)

\(\Rightarrow A⋮4\left(ĐPCM\right)\)

Ta có ; \(A=3+3^2+3^3+.....+3^{100}\)

                \(=\left(3+3^2+3^3+3^4+3^5\right)\)

dễ mà bạn bạn cứ nhóm 3số đầu tiên vào roi cu tiep tuc 3 so nhu vay

se duoc : (1+3+3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)

=(1+3+3^2)+3^3.(1+3+3^2)+...+3 ^98.(1+3+3^2)

=13.3^3.13+...+3^98.13=13.(1+3^3+...+3^98) chia hết cho 13 

vậy M chia hết cho 13

tick cho mình nhé!