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\(A=1+6+6^2+...+6^{100}\)
\(6A=6+6^2+6^3+...+6^{101}\)
\(6A-A=\left(6+6^2+...+6^{101}\right)-\left(1+6+...+6^{100}\right)\)
\(5A=6^{101}-1\)
\(A=\frac{6^{101}-1}{5}\)
Hoàn toàn tương tự với các câu b) c)
\(A=1+6+6^2+6^3+...+6^{100}\)
\(6A=6+6^2+6^3+6^4+...+6^{101}\)
\(6A-A=\left(6+6^2+6^3+6^4+...+6^{101}\right)-\left(1+6+6^2+...+6^{100}\right)\)
\(5A=6^{101}-1\)
\(A=\frac{6^{101}-1}{5}\)
a, A = 1 + 2 + 22 + ... + 299
= (1 + 2) + (22 + 23) + ... + (298 + 299)
= 1(1 + 2) + 22(1 + 2) + ... + 298(1 + 2)
= 1 . 3 + 22 . 3 + ... + 298 . 3
Vì 3 chia hết cho 3 nên 1 . 3 + 22 . 3 + ... + 298 . 3 chia hết cho 3
hay A chia hết cho 3 (đpcm)
b, A = 1 + 2 + 22 + ... + 299
= (1 + 2 + 22 + 23) + (24 + 25 + 26 + 27) + ... + (296 + 297 + 298 + 299)
= 1 . 15 + 24 . 15 + ... + 296 . 15
Vì 15 chia hết cho 15 nên 1 . 15 + 24 . 15 + ... + 296 . 15 chia hết cho 15
hay A chia hết cho 15 (đpcm)
Tiếp bài của @trankhanhvy2008
A = 1 + 2 + 22 + 23 + 24 + ... + 299
2A = 2( 1 + 2 + 22 + 23 + 24 + ... + 299 )
= 2 + 22 + 23 + 24 + ... + 2100
2A - A = ( 2 + 22 + 23 + 24 + ... + 2100 ) - ( 1 + 2 + 22 + 23 + 24 + ... + 299 )
=> A = 2 + 22 + 23 + 24 + ... + 2100 - 1 - 2 - 22 - 23 - 24 - ... - 299
= 2100 - 1
2100 - 1 < 2100
=> A < 2100
số sh cua tong A bang so hang cua day so cach deu 1 don vi tu 1 den 60
so sh cua tong A la:(60-1):1+1=60 (sh)
Cu 3 sh lien tiep cua tong A nhom thanh 1 nhom thi ta duoc so nhom la : 60: 3=20(nhom)
khi do : A = (2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)+....+(2^58+2^59+2^60)
A=(2+2.2+2.2^2)+(2^4+2^4.2+2^4.2^2)+(2^7+2^7.2+2^7.2^2)+.....+(2^58
2^58.2+2^58.2^2)
A=2(1+2+2^2)+2^4(1+2+2^2)+2^7(1+2+2^2)+...+2^58(1+2+2^2)
A=2.7+2^4.7+2^7.7+...+2^58.7
A=7(2+2^4+2^7+...+2^58)
Vi 7 chia het cho 7
2+2^4+2^7+...+2^58 thuoc N
Suy ra 7(2+2^4+2^7+...+2^58) chia het cho 7
hay A chia het cho 7
Vay A chia het cho 7
Câu 1:
abc >/ 100 ; bca >/ 100 ; cab>/100
< = > abc + bca + cab >/300
< = > abc + bca + cab >/ 111
2A=2+22+23+...+2101
2A+1=1+2+22+...+2101=A+2101
2A-A=2101-1
A=2101-1
nên 250*(A+1)=250*(2101-1+1)=250*2101=2151
Vậy m=151
Ban dich day :
Cho A=1+2+22+23+...+2100
Neu 250. (A+1)=2m thi m=
Tra loi: m=
Tich nha!
1) Đặt \(A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{99}.3\)
Vì \(3⋮3\) nên \(2.3+2^3.3+...+2^{99}.3⋮3\)
hay \(A⋮3\)(đpcm)
2) Đặt \(B=3+3^2+3^3+...+3^{1998}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{1996}+3^{1997}+3^{1998}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{1996}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{1996}.13\)
\(=39+3^3.39+...+3^{1995}.39\)
Vì \(39⋮39\)nên \(39+3^3.39+...+3^{1995}.39⋮39\)
hay \(B⋮39\)(đpcm)
a) 2+22+23+...+2100
=(2+22+23+24+25)+(26+27+28+29+210)+.....+(296+297+298+299+2100)
=2(1+2+22+23+24)+26(1+2+22+23+24)+....+296(1+2+22+23+24)
=2(1+2+4+8+16)+26(1+2+4+8+16)+....+296(1+2+4+8+16)
=2.31+26.31+....+296.31
=31(2+26+....+296)
=> đpcm