Sửa đề: \(B=2+2^2+2^3+...+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)⋮5\)

\(B=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\left(2+2^6+...+2^{96}\right)⋮31\)

did you studied at le van tam primary school

giúp với mình đang cần gấp

Sao biết tớ

Gọi C là giá trị của biểu thức trên

a) CMR : C chia hết cho 31

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

\(C=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{19}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(C=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(C=2.31+2^6.31+...+2^{96}.31\)

\(C=31\left(2+2^6+2^{10}+...+2^{96}\right)⋮31\)(đpcm) 

b) CMR : C chia hết cho 5

\(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{97}+2^{99}\right)+\left(2^{98}+2^{100}\right)\)

\(=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{97}\left(1+2^2\right)+2^{98}\left(1+2^2\right)\)

=\(2.5+2^2.5+...+2^{97}.5+2^{98}.5\)

\(=5\left(2+2^2+...+2^{97}+2^{98}\right)⋮5\)(đpcm)

Vậy 2 + 2^2 + 2^3 + ...+ 2^98 + 2^99 + 2^100 vừa chia hết cho 5 vừa chia hết cho 31

\(C=2+2^2+2^3+......+2^{100}⋮31\)

\(C=2.\left(1+2+2^2+2^3+2^4\right)+2^{95}.\left(1+2+2^2+2^3+2^4\right)\)

\(C=2.31+.......+2^{95}+31\)

\(C=31.\left(2+2^{95}\right)⋮31\)

\(\Rightarrow C⋮31\)

c=2^101-2 chia hết cho 31

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=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)

bạn lấy \(2^{200}\)trừ cho số cuối rồi đóng ngoặc lại sau đó cộng với một bạn chỉ lấy vài số để trừ thồi nha

chúc bạn học tốt