chứng minh: K=1/3+2/3^2+...+2008/3^2008 <3/4
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VV
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VC
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20 tháng 12 2017
1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)
\(\dfrac{A}{B}=\dfrac{1}{2009}\)
2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
LM
29 tháng 10 2017
Đặt A =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}\)
Suy ra 3A = \(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2008}{3^{2007}}\)=> 2A = 3A - A = \(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2008}{3^{2007}}-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{2008}{3^{3008}}\)= \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}-\frac{2008}{3^{2008}}\)
= \(\frac{3}{2}-\frac{1}{2.3^{2007}}\)Suy ra A = \(\frac{3}{4}-\frac{1}{8.3^{2007}}\)<\(\frac{3}{4}\)(ĐPCM)
2 tháng 6 2015
Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:
\(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)
\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)
\(=2009^{2008}-1\)
\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)
\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008
=> ĐPCM
VT
2 tháng 6 2015
Chứng Minh Rằng: (2009+20092+20093+20094+...+20092009)-(1+2009+20092+20093+...+20092008) chia hết cho 2008.
Đặt A=2009+20092+20093+20094+...+20092009, B=1+2009+20092+20093+20094+...+20092008
Ta có:
+)A=2009+20092+20093+20094+...+20092009
2009A= 20092+20093+20094+...+20092010
2009A-A=(20092+20093+20094+...+20092010)-(2009+20092+20093+20094+...+20092009)
2008A=20092010- 2009
=> A=(20092010- 2009)/2008
=> A chia hết cho 2008.
B=1+2009+20092+20093+20094+...+20092008
2009B=2009+20092+20093+20094+...+20092010
2009B-B=(2009+20092+20093+20094+...+20092010)-(1+2009+20092+20093+20094+...+20092009)
2008B=20092010-1
=>B=(20092010-1)/2008
=>B chia hết cho 2008
=> A-B chia hết cho 2008.
=> ĐPCM
3 tháng 10 2015
1)A=3+32+33+...+32008
A=(3+32)+(33+34)+...+(32007+32008)
A=3(1+3)+33(1+3)+...+32007(1+3)
A=3.4+33.4+...+32007.4
A=4(3+....+32007) chia hết cho 4
ZS
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