Ta có : S = 2 + 22 + 23 + 24 + 25 + 26 + ... + 297 + 298 + 299
= (2 + 22 + 23) + (24 + 25 + 26) + ... + (297 + 298 + 299)
= (2 + 22 + 23) + 23. (2 + 22 + 23) + ... + 296. (2 + 22 + 23)
= 14 + 23.14 + ... + 296.14
= 14.(1 + 23 + ... + 296) \(⋮\)14
=> \(S⋮14\left(\text{ĐPCM}\right)\)
Ta có : S=2+22+23+...+299
=(2+22+23)+(24+25+26)+...+(297+298+299)
=2(1+2+22)+24(1+2+22)+...+297(1+2+22)
=2.7+24.7+...+297.7
=14+23.2.7+...+296.2.7
=14.23.14+...+296.14
Vì 14\(⋮\)14 nên 14.23.14+...+296.14\(⋮\)14
hay S\(⋮\)14
Vậy S\(⋮\)14.
S = 2 + 22 + 23 + ... + 298 + 299
= ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ...+ ( 297 + 298 + 299 )
= 1 . 14 + 23 . 14 + ... + 296 . 14
= 14 . ( 1 + 23 + ... + 296 ) Chia hết cho 14
HỌC TỐT !
S = 2+2^2+2^3+...+2^98+2^99.
=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^97+2^98+2^99)
=1.(1+2^2+2^3)+2^2.(1+2^2+2^3)+...+2^96.(1+2^2+2^3)
=1.14+2^2.14+...+2^96.14
=14. (1+2^2+2^3+...+2^96) chia hết cho 14(điều ta cần chứng ming)