cho S=3+3233+...+3100
chứng minh rằng S chia hết cho 40
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10 tháng 2 2019
\(A,\)\(S=\left(3+3^2\right)+\left(3+3^2\right)3^2+...+\left(3+3^2\right)3^{2018} \)
\(\Rightarrow S=9\left(1+3^2+...+3^{2018}\right)\)
\(\Rightarrow S⋮9\)
\(B,\)\(S=3+3^2+3^3+\left(3+3^2+3^3\right)3^3+...\left(3+3^2+3^3\right)3^{2017}\)
\(S=39+39.3^3+...+39.3^{2017}\)
Nhưng xét lại thì thấy 2017 không chia hết cho 3 nên câu b có lẽ sai đề =)))))
\(C,\)\(S=\left(1+3+3^2+3^3\right).3+\left(1+3+3^2+3^3\right).3^4+...+\left(1+3+3^2+3^3\right).3^{2017}\)
\(S=40.3+40.3^4+...+40.3^{2017}\)
\(Vậy...\)
LN
1
AH
Akai Haruma
Giáo viên
16 tháng 7
Lời giải:
$S=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{47}+3^{48}+3^{49}+3^{50})$
$=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+.....+3^{47}(1+3+3^2+3^3)$
$=(1+3+3^2+3^3)(3+3^5+...+3^{47})=40(3+3^5+...+3^{47})\vdots 40$
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
1 tháng 8 2023
Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
ND
Nguyễn Đức Trí
VIP
1 tháng 8 2023
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
TT
1
IW
4 tháng 10 2016
a) \(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+.....+\left(3^{88}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+......+3^{88}\left(1+3\right)\)
\(\Rightarrow A=1.4+3^2.4+..........+3^{88}.4\)
\(\Rightarrow A=4.\left(1+3^2+.........+3^{88}\right)\)
Vậy A chia hết cho 4 ĐPCM
b) \(\Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)\)\(+......+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+\)\(....+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=1.40+3^4.40+.......+3^{96}.40\)
\(\Rightarrow A=40.\left(1+3^4+....+3^{96}\right)\)
Vậy A chia hết cho 40 ĐPCM
30 tháng 6 2016
B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
6 tháng 9 2018
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
\(S=3+3^2+3^3+...+3^{100}\)
\(S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(S=40.3+...+3^{96}\left(3+3^2+3^3+3^4\right)\)
\(S=40.3+...+3^{96}.40.3\)
\(S=40.3.\left(3^4+...+3^{96}\right)\)chia hết 40
Ta có: S = 3 + 32 + 33 + ...... + 3100
=> 3S = 32 + 33 + 33 +...... + 3101
=> 3S - S = 3101 - 3
=> 2S = 3101 - 3
=> S = \(\frac{3^{101}-3}{2}\)