\(a,S=\dfrac{\left(2014+4\right)\left[\left(2014-4\right):3+1\right]}{2}=\dfrac{2018\cdot671}{2}=677039\\ b,\forall n\text{ lẻ }\Rightarrow n+2013\text{ chẵn }\Rightarrow n\left(n+2013\right)⋮2\left(1\right)\\ \forall n\text{ chẵn }\Rightarrow n\left(n+2013\right)⋮2\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\\ c,M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{10}\right)\\ M=2\left(1+2+2^2+2^3\right)+...+2^{16}\left(1+2+2^2+2^3\right)\\ M=\left(1+2+2^2+2^3\right)\left(2+...+2^{16}\right)=15\left(2+...+2^{16}\right)⋮15\)

Đáp án đây nhé : https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++Ch%E1%BB%A9ng+t%E1%BB%8F+M+chia+h%E1%BA%BFt+cho+5M=2+22+23+...+220&id=733479

Bạn chép vào đi 

M = (2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^17+2^18+2^19+2^20)

    = 2.(1+2+2^2+2^3)+2^5.(1+2+2^2+2^3)+.....+2^17.(1+2+2^2+2^3)

    = 2.15 + 2^5.15 +.... +2^17.15

    = 15.(2+2^5+....+2^17) chia hết cho 15

co cai nit tu di ma tinh

Giải : A = 2 + 2² + 2³ + ........ + 2²⁰

          2A = 4 + 2³ + 2⁴ + ........ + 2²¹

Suy ra : 2A - A = 2²¹ + 4 - ( 2 + 2² )

Vậy    :        A = 2²¹

ở trường thì bn hok đi lại còn vào lm zè!

M=2+2²+2³+...+2²⁰

=> M= (2+2²+2³+2⁴)+...+(2¹⁷+2¹⁸+2¹⁹+2²⁰)

=> M = 2.(1+2+2²+2³)+...+2¹⁷.(1+2+2²+2³)

=>  M= 2.15+...+2¹⁷.15

=> M = 15.(2+...+2¹⁷) = 3.5.(2+...+2¹⁷) chia het cho 5 

=> M chia hết cho 5 

\(\Leftrightarrow M=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(\Leftrightarrow M=30+2^4\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow M=30+2^4.30+...+2^{16}.30\)

\(\Leftrightarrow M=30\left(1+2^4+...+2^{16}\right)⋮5\)

\(M=\left(2+2^2+2^3+2^4\right)+...+2^{17}\left(2+2^2+2^3+2^4\right)\)

\(=30\cdot\left(1+...+2^{17}\right)⋮5\)

⇔M=(2+22+23+24)+(25+26+27+28)+...+(217+218+219+220)

⇔M=30+24(2+22+23+24)+...+216(2+22+23+24)

⇔M=30+24.30+...+216.30

⇔M=30(1+24+...+216)⋮5

\(M=2+2^2+2^3+2^4+...+2^{20}\\ =\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{18}\left(2+2^2\right)\\ =6+2^2.6+...+2^{18}.6\\ =\left(1+2^2+...+2^{18}\right).6⋮6\)

M = 2 + 2² + 2³ + ... + 2²⁰

M = 2¹ + 2² + 2³ + ... + 2²⁰

Xét dãy số: 1; 2; 3;...; 20 dãy số này có 20 số hạng vậy M có 20 hạng tử. Vì 20 : 2 = 10 nên nhóm 2 hạng tử liên tiếp của M thành 1 nhóm thì:

M = (2¹ + 2²) + (2³ + 2⁴) + ... + (2¹⁹ + 2²⁰)

M = 6 + 2².( 2+ 2²) + ... + 2¹⁸(2 + 2²)

M = 6 + 2².6 + ... + 2¹⁸. 6

M = 6. ( 1 + 2² + ... + 2¹⁸)

vì 6 ⋮ 6 nên 6.(1 + 2² + ... + 2¹⁸) ⋮ 6 hay M = 2 + 2²+...+2²⁰ ⋮ 6(đpcm)

kinh

A=4+(2²+2³+2⁴+...+2²⁰)

                  A-4=2²+2³+2⁴+...+2²⁰

               2(A-4)=2³+2⁴+2⁵+...+2²¹

A-4=2(A-4)-(A-4)=(2³+2⁴+2⁵+...+2²¹)-(2²+2³+2⁴+...+2²⁰)

                   A-4=(2³-2³)+(2⁴-2⁴)+(2⁵-2⁵)+...+(2²⁰-2²⁰)+(2²¹-2²)

                   A-4=2²¹-4

                   A   =2²¹-4+4

                   A   =2²¹

\(M=2\left(1+2+2^2+...+2^{19}\right)⋮2\)

\(M=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)=\)

\(=3\left(2+2^3+2^5+...2^{19}\right)⋮3\)

\(M=\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{17}+2^{19}\right)+\left(2^2+2^4\right)+...+\left(2^{18}+2^{20}\right)\)

\(M=2\left(1+2^2\right)+2^5\left(1+2^2\right)+...+2^{17}\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)

\(M=2.5+2^5.5+...+2^{17}.5+...+2^{18}.5⋮5\)