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De thay B co 996 so hang
Ta co: 3+3^3+3^5+...+3^1991
= (3+3^3+3^5)+...+(3^1987+1989+1991)
=3.(1+3^2+3^4)+...+3^1987.(1+3^2+3^4)
=3.91+...+3^1987.91
=(3+..+3^1987).91=(3+...+3^1987).13.7 chia het cho 13
3+3^3+3^5+...+3^1991
=(3+3^3+3^5+3^7)+...+(3^1985+3^1987+3^1989+3^1991)
=3(1+3^2+3^4+3^6)+...+3^1985.(1+3^2+3^4+3^6)
=3.820+...+3^1985.820=(3+...+3^1985).820=(3+....+3^1985).41.20 chia het cho 41
chưng tỏ B:13
B=3+33+35+...+31991:13
B=3. (1+9+81)+37.(1+9+81)+...+31989.(1+9+81):13
B=91.(3+37+313+...+31989):13
vì 91:13=>B:13
vậy B:13
chưng tỏ B:41
B=3+33+35+...+31991:41
B=3.(1+9+81+729)+39.(1+9+81+729)+...+31988.(1+9+81+729):41
B=820.(3+39+317+...+31988):41
vì 820:41=>B:41
vậy B:41
a) A= (2+22)+(23+24)+........(259+260)
= 1(2+22) + 22(2+22) + ....... 258(2+22)
= 1.6 + 22.6 +......... 258.6
=6(1+22+.......258)
Vì 6 chia hết cho 3 nên => 6(1+22+........258)
Các câu còn lại cũng tương tự như vậy nha bn!
Ta có: `B = 1 + 3 + 3^2 + ... + 3^1991`
`= (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^1989 + 3^1990 + 3^1992)`
`= 13 + 3^3 (1 + 3 + 3^2) + ... + 3^1989 (1 + 3 + 3^2)`
`= 13 + 3^3 . 13 + ... + 3^1989 . 13`
`= 13 (1 + 3^3 + ... + 3^1989)`
Vì \(13\left(1+3^3+...+3^{1989}\right)⋮13\) nên \(B⋮13\)
`B = 1 + 3 + 3^2 + ... + 3^1991`
= (1 + 3^4) + (3 + 3^5) + ... + (3^1987 + 3^1991)`
`= 82 + 3 (1 + 3^4) + ... + 3^1987 (1 + 3^4)`
`= 82 + 3 . 82 + ... + 3^1987 . 82`
`= 82 (1 + 3 + ... + 3^1987)`
Vì \(82\left(1+3+...+3^{1987}\right)⋮41\) nên \(B⋮41\)
`C = 3 + 3^2 + 3^3 + ... + 3^1000`
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{997}+3^{998}+3^{999}+3^{1000}\right)\)
`= 120 + 3^4 (3 + 3^2 + 3^3 + 3^4) + ... + 3^996 (3 + 3^2 + 3^3 + 3^4)`
`= 120 + 3^4 . 120 + ... + 3^996 . 120`
`= 120 (1 + 3^4 + ... + 3^996)`
Vì \(120\left(1+3^4+...+3^{996}\right)⋮120\) nên \(C⋮120\)
Ta có: \(C=3+3^2+3^3+...+3^{1000}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{997}+3^{998}+3^{999}+3^{1000}\right)\)
\(=120\left(1+3^5+...+3^{997}\right)⋮120\)(đpcm)