Cho B = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^2004 + 1/3^2005Chứng minh rằng B <...
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2 tháng 6 2015
a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004
B= 1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005
suy ra 2B=1-1/3^2005
suy ra B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
suy ra B=1/2-1/3^2005/2 bé hơn 1/2
từ đấy suy ra B bé hơn 1/2
LA
11 tháng 2 2018
Từng bài 1 thôi nhs!
a) 3A = 3 - 32 + 33 - 34 + ... -32004+ 32005
3A + A = 3 - 32 + 33 -34 + ... -32004 + 32005 +1 - 3 + 32- 33 + 34 - ....-32003+32004
4A = 32005 + 1
=> 4A - 1 = 32005 là lũy thừa của 3
=> ĐPCM
TT
14 tháng 6
đề có thiếu ko đó
A = 4 + 23 + 24 + 25 + ...+ 22003 + 22004
đặt B = 23 + 24 + 25 + ...+ 22003 + 22004
2B= 24 + 25 + 26 + ....+ 22004 + 22005
2B-B= ( 24 + 25 + 26 + ....+ 22004 + 22005 ) - ( 23 + 24 + 25 + ...+ 22003 + 22004 )
B = 24 + 25 + 26 + ....+ 22004 + 22005 - 23 - 24 - 25 - ...- 22003 - 22004
B = 22005 - 23
B = 22005 - 8
=> A = 4 + B = 4 + 22005 - 8 = 22005 - 4 = .....
VT
1
14 tháng 1 2016
Ta có:3B\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+...+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}\)
B=\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+..+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}+\frac{1}{3}^{2005}\)
\(\Rightarrow\)2B=1-\(\frac{1}{3}^{2005}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}<\frac{1}{2}\)
\(\Rightarrow\)B<\(\frac{1}{2}\)
3 tháng 3 2016
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
LC
3 tháng 3 2016
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005
B=1/2-1/(32005.2)
Vậy B <1/2
VH
1
27 tháng 4 2017
Ta có :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow\frac{2B}{2}=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
LH
3
6 tháng 3 2017
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=3\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\) (Đpcm)
6 tháng 3 2017
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ \)
\(C=3B=1+\dfrac{1}{3}+..+\dfrac{1}{3^{2004}}\)
\(C-B=1-\dfrac{1}{3^{3005}}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2.3^{2005}}< \dfrac{1}{2}\)
8 tháng 8 2016
Bài 1
a) 34 + 35 + 36 + 37 = 34(1 + 3 + 32 + 33)\
b) a)A = 1 + 3 + 32 +......399 =(1 + 3 + 32 + 33 ) + ...+(396 + 397 + 398 + 399)
= (1 + 3 + 32 + 33 ) + .. +396(1 + 3 + 32 + 33 )
= 40 + ... + 396 . 40
= 40 (1 + 3 +...+ 396) chia hết cho 40
8 tháng 8 2016
Bài 2
a)
+)A chia hết cho 6
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{2002}\left(5+5^2\right)\)
\(A=30+5^2.30+...+5^{2002}.30\)
\(A=30\left(1+5^2+...+5^{2002}\right)\)chia hết cho 6
+)A chia hết cho 31
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{2002}+5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2+5^3\right)+5^3\left(5+5^2+5^3\right)+...+5^{2001}\left(5+5^2+5^3\right)\)
\(A=155+5^3.155+...+5^{2001}.155\)
\(A=155\left(1+5^3+...+5^{2001}\right)\)chia hết cho 31
+) A chia hết cho 156
\(A=5+5^2+5^3+...+5^{2004}\)
\(A=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)
\(A=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2000}\left(5+5^2+5^3+5^4\right)\)
\(A=780+5^4.780+...+5^{2000}.780\)
\(A=780\left(1+5^4+...+5^{2000}\right)\)chia hết cho 156
b)B=165+2^15 chia hết cho 33
ta có 165 chia hết cho 33
mà 215 ko chia hết cho 33
vậy 165+2^15 không chia hết cho 33 hay B không chia hết cho 33.
QQ
0
3B=1+1/3+...+1/3^2004
=>2B=1-1/3^2005
=>\(2B=\dfrac{3^{2005}-1}{3^{2005}}\)
=>\(B=\dfrac{3^{2005}-1}{3^{2005}\cdot2}< \dfrac{1}{2}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +........+ \(\dfrac{1}{3^{2024}}\)+ \(\dfrac{1}{3^{2005}}\)
3B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +........+\(\dfrac{1}{3^{2004}}\)
3B -B = 1 - \(\dfrac{1}{3^{2005}}\)
2B = 1 - \(\dfrac{1}{3^{2005}}\)
B = ( 1 - \(\dfrac{1}{3^{2005}}\)):2
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{2005}}\) < \(\dfrac{1}{2}\) (đpcm)