Cho A =2+2^2+2^3+...+2^60 chứng tỏ rằng A chia hết cho 3,5
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T
1 tháng 12 2023
a) \(A=2+2^2+2^3+\dots+2^{60}\)
\(2A=2^2+2^3+2^4+\dots+2^{61}\)
\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)
\(A=2^{61}-2\)
Vậy: \(A=2^{61}-2\).
b)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)
\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)
\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)
Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)
\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)
\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)
Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)
hay \(A\vdots5\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)
\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)
\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)
Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)
$Toru$
1L
1
NV
0
CC
2
9 tháng 7 2017
TA có:VÌ 2= 2^1
A=\(2^1+2^2+2^3+...+2^{60}\)
A= \(\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
A= \(2\left(1+2\right)+2^3\left(1+2\right)+...2^{59}\left(1+2\right)\)
A= \(3.\left(2+2^3+...+2^{60}\right)\)chia hết cho 3
=) A chia hết cho3( đpcm)
Ta lại có:
A= \(2^1+2^2+2^3+...+2^{60}\)
A= \(\left(2^1+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
A=\(2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
A= \(7.\left(2+...+2^{58}\right)\)chia hết cho 7
=) A chia hết cho 7( đpcm)
28 tháng 4 2015
A= (21+22+23)+(24+25+26)+...+(258+259+260)
=20(21+22+23)+23(21+22+23)+...+257(21+22+23)
=(21+22+23)(20+23+...+257)
= 14(20+23+...+257) chia hết cho 7
Vậy A chia hết cho 7
25 tháng 6 2015
gọi 1/41+1/42+1/43+...+1/80=S
ta có :
S>1/60+1/60+1/60+...+1/60
S>1/60 x 40
S>8/12>7/12
Vậy S>7/12
H9
HT.Phong (9A5)
CTVHS
21 tháng 10 2023
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
NM
0
TT
2
LM
26 tháng 7 2016
A=2+22+23+....+260
A=(2+22)+(23+24)+...+(259+260)
A=2.3+23.3+....+259.3 chia hết cho 3
2) A=2+22+23+...+260
A=(2+22+23)+.... +(258+259+260)
A=2.7+....+258.7 chia hết cho 7
3) A=2+22+23+....+260
A=(2+22+23+24)+....+(257+258+259+260)
A=2.15+....+257.15 chia hết cho 15
26 tháng 7 2016
A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3
=>A chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7 chia hết cho 7 =>7.(2+...+258) chia hết cho 7
=>A chia hết cho 7
A= (2+22+23+24)+...+(257+258+259+260)
A=2.(1+2+22+23)+...+257.(1+2+22+23)
A=2.15 +...+257.15
A=15.(2+...+257)
vì 15 chia hết cho15=>15.(2+...+25) chia hết cho 15
=>A chia hết cho 15
\(2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\\ =2.15+2^5.15+...+2^{57}.15=15\left(2+2^5+...+2^{57}\right)\)
Mà \(15\left(2+2^5+...+2^{57}\right)⋮3\) và \(15\left(2+2^5+...+2^{57}\right)⋮5\) nên A chia hết cho 3 và 5