2+2²+...+2¹⁰⁰

=(2+2²)+...+(2⁹⁹+2¹⁰⁰)

=2(1+2)+...+2⁹⁹(1+2)

=3.(2+...+2⁹⁹) chia hết cho 3

Vậy 2+2²+...+2¹⁰⁰ chia hết cho 3

chắc bạn chép sai đầu bài ý a rồi , mình sửa lại nhé

Đặt A=\(2+2^2+2^3+...+2^{100}\)

Tổng A có :(100-1):1+1=100(số hạng)

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

A=\(\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

(có \(\dfrac{100}{5}=20\) nhóm , mỗi nhóm có 5 số hạng)

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

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

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

Sửa đề câu a tí nhé:

Chứng tỏ \(\left(2+2^2+2^3+...+2^{100}\right)\)chia hết cho 31

Giải:

Đặt \(S=\left(2+2^2+2^3+...+2^{100}\right)\)

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

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

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

\(\Rightarrow S⋮31\)

b.

Đặt \(A=\left(1+3+3^2+3^3+...+3^{11}\right)\)

\(A=\left(1+3+3^2+3^3+...+3^8.\left(1+3+3^2+3^3\right)\right)\)

\(A=40+...+3^8.40\)

\(A=40.\left(1+...+3^8\right)⋮40\)

Vậy \(A\) chia hết cho \(40\)

TA CÓ:

A=3⁰+3+3²+3³+........+3¹¹

(3⁰+3+3²+3³)+....+(3⁸+3⁹+3¹⁰+3¹¹)

3(0+1+3+3²)+......+3⁸(0+1+3+3²) 

3.13+....+3⁸.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)

co cai nit tu di ma tinh

1) \(1+4+4^2+4^3+...+4^{2012}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{2010}+4^{2011}+4^{2012}\right)\)

\(=21+21\cdot4^3+...+21\cdot4^{2010}\)

\(=21\cdot\left(1+4^3+...+4^{2010}\right)\) chia hết cho 21

2) \(1+7+7^2+7^3+...+7^{101}\)

\(=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{100}+7^{101}\right)\)

\(=8+8\cdot7^2+...8\cdot7^{100}\)

\(=8\cdot\left(1+7^2+...+7^{100}\right)\) chia hết cho 8

3) CM chia hết cho 5:

\(2+2^2+2^3+2^4+...+2^{100}\)

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

\(=5\cdot2+5\cdot2^2+...+5\cdot2^{98}\)

\(=5\cdot\left(2+2^2+...+2^{98}\right)\) chia hết cho 5

CM chia hết cho 31:

\(2+2^2+2^3+...+2^{100}\)

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

\(=2\cdot31+...+2^{96}\cdot31\)

\(=31\cdot\left(2+...+2^{96}\right)\) chia hết cho 31

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3

Tổng trên sẽ bằng:

 \(\left(70000+1\right).70000:2=70001.35000⋮100\)

Vậy ta có đpcm.

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

=12(1+3^2+...+3^98) chia hết cho 12

gọi A = 2+2^2+2^3+......+2^100

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+......+(2^97+2^98+2^99+2^100)

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+......+(2^97+2^98+2^99+2^100)

A=(2+2^2+2^3+2^4).1+(2+2^2+2^3+2^4).4+......+(2+2^2+2^3+2^4).98

A=            30.1          +         30.4             +.......+       30.98

A=                  30.(1+4+...+98)

Vì 30 chia hết cho 3

=>30.(1+4+...98) chia hết cho 3

Hay 2+2^2+2^3+......+2^100 chia hết cho 3.