chứng tỏ\(A=3+3^2+3^3+...+3^{59}+3^{60}⋮13\)
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SM
3
HT
6 tháng 11 2020
ta có :
A = 3 + 32 + 33 + ...+ 359 + 360
A = ( 3 + 32 + 33 ) + ( 34 + 35 + 36 ) + ...+ ( 358 + 359 + 360 )
A = 3( 1 + 3 + 32) + 34(1+3+32) + ...+ 358(1+3+32 )
A = 3. 13 + 34.13 + ...+ 358.13
=> A chia hết cho 13
NM
Nguyễn Minh Quang
Giáo viên
6 tháng 11 2020
Ta chú ý : \(3+3^2+3^3=3\left(1+3+9\right)=3.13\)
\(\Rightarrow A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+..+\left(3^{58}+3^{59}+3^{60}\right)\)
\(\Leftrightarrow A=3.13+3^4.13+...+3^{58}.13\)
\(\Leftrightarrow A=13\left(3+3^4+..+3^{58}\right)⋮13\)
Vậy A chia hết cho 13
ND
1
KT
1 tháng 8 2018
\(A=3+3^2+3^3+....+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+....+3^{59}\right)\)
\(=4\left(3+3^3+....+3^{59}\right)\)\(⋮\)\(4\)
\(A=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(3+3^4+....+3^{58}\right)\)
\(=13\left(3+3^4+...+3^{58}\right)\)\(⋮\)\(13\)
mà (4;13) = 1
nên A chia hết cho 52
PK
1
H9
HT.Phong (9A5)
CTVHS
30 tháng 10 2023
\(A=3^1+3^2+...+3^{60}\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(A=3\cdot\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)
\(A=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{58}\cdot\left(1+3+9\right)\)
\(A=3\cdot13+3^4\cdot13+...+3^{58}\cdot13\)
\(A=13\cdot\left(3+3^4+...+3^{58}\right)\)
Mà: \(13\cdot\left(3+3^4+...+3^{58}\right)\) ⋮ 13
\(\Rightarrow A\) ⋮ 13
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 9 2016
\(A=\left(3^{60}+3^{58}+3^{56}+...+3^2\right)-\left(3^{59}+3^{57}+3^{55}+...+3\right).\)
\(B=3^{60}+3^{58}+3^{56}+...+3^2\)
\(9B=3^{62}+3^{60}+3^{58}+...+3^4\)
\(B=\frac{9B-B}{8}=\frac{3^{62}-3^2}{8}=\frac{3^2\left(3^{60}-1\right)}{8}\)
\(C=3^{59}+3^{57}+3^{55}+...+3\)
\(9C=3^{61}+3^{59}+3^{57}+...+3^3\)
\(C=\frac{9C-C}{8}=\frac{3^{61}-3}{8}=\frac{3\left(3^{60}-1\right)}{8}\)
\(A=B-C=\frac{3^2\left(3^{60}-1\right)-3\left(3^{60}-1\right)}{8}=\frac{6\left(3^{60}-1\right)}{8}\)
\(A=\frac{2.3.\left(3^{60}-1\right)}{8}=\frac{2.3.3^{60}}{8}-\frac{2.3}{8}=\frac{3^{61}}{4}-\frac{3}{4}=\frac{3^{61}-3}{4}\)
CN
2
MH
16 tháng 12 2015
Đặt tổng trên là A
Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{59}.\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{59}.3\)
\(=3.\left(2+2^3+...+2^{59}\right)\)chia hết cho 3
=> A chia hết cho 3 (Đpcm).
16 tháng 12 2015
Ta có :
2+2^2+2^3+2^4+...+2^59+2^60=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)
=2x3+2^3x3+...+2^59x3
=(2+2^3+...+2^59)x3
Vì 3 chia hết cho 3 nên tổng trên chia chiết cho 3 (đpcm)
TP
0
20 tháng 12 2021
b: \(A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{58}\right)⋮13\)
20 tháng 12 2021
\(a,\Leftrightarrow2A=8+2^3+2^4+...+2^{21}\\ \Leftrightarrow2A-A=8+2^3+2^4+...+2^{21}-4-2^2-2^3-...-2^{20}\\ \Leftrightarrow A=2^{21}+8-4-2^2=2^{21}\left(đpcm\right)\\ b,A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)\\ A=13\left(3+3^4+...+3^{58}\right)⋮13\)
Ta có :
\(A=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(3+3^4+3^7+...+3^{60}\right)\)
\(=13\cdot\left(3+3^4+3^7+...+3^{60}\right)⋮13\)(ĐPCM)
Ta có : A=3+32+33+...+360
=(3+32+33)+(34+35+36)+...+(358+359+360)
=3(1+3+32)+34(1+3+32)+...+358(1+3+32)
=3.13+34.13+...+358.13
Vì 13 chia hết cho 13 nên 3.13+34.13+...+358.13 chia hết cho 13
=> A chia hết cho 13(đpcm)