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đặt \(A=1+2+2^2+2^3+...+2^{2008}=2-1+2^2-2+2^3-2^2+...+2^{2009}-2^{2008}\)
\(=\left(2-2\right)+\left(2^2-2^2\right)+\left(2^3-2^3\right)+...+2^{2009}-1=2^{2009}-1\)
\(2^{2009}-1+1-2^{2009}=0\)
=>A và \(1-2^{2009}\)đối nhau
\(\Rightarrow B=\frac{A}{-A}=-1\)
\(2B=\frac{2+2^2+2^3+2^4+...+2^{2009}}{1-2^{2009}}\)
\(B=2B-B=\frac{\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)}{1-2^{2009}}\)
\(\Leftrightarrow B=\frac{2^{2009}-1}{1-2^{2009}}\)
Vì 1<22009 nên 22009-1 là số đối của 1-22009 nên suy ra B = -1
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )
A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 )
A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )
A = ( - 4 ) . 502 + 4019
A = - 2008 + 4019
A = 2011
\(A=1+2-3-4+5+6-7-8+...+2005+2006-2007-2008\)\(+2009+2010\)
( Có 2010 số hạng )
\(A=\left(1+2-3-4\right)+.....+\left(2005+2006-2007-2008\right)+2009+2010\)
( Có 502 nhóm )
\(A=\left(-4\right)+\left(-4\right)+......+\left(-4\right)+2009+2010\)
( Có 502 số - 4 )
\(A=-4\cdot502+2009+2010\)
\(A=-2008+2009+2010\)
\(A=1+2010\)
\(A=2011\)
Ta có \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2021}\right)\left(1-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}.\dfrac{2021}{2022}\)
\(B=\dfrac{1}{2022}\)
1)Đặt A=1+2+22+23+.....+22008
=>2A=2+22+23+....+22009
=>2A-A=(2+22+23+...+22009)-(1+2+22+23+....+22008)
=-1+22009
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}=\dfrac{1}{2022}\)
\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(1-\dfrac{1}{2021}\right)\cdot\left(1-\dfrac{1}{2022}\right)\)
\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\cdot\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(\dfrac{2021}{2021}-\dfrac{1}{2021}\right)\cdot\left(\dfrac{2022}{2022}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}\)
\(B=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot2020\cdot2021}{2\cdot3\cdot4\cdot\cdot\cdot2021\cdot2022}\)
\(B=\dfrac{1}{2022}\)
nhóm như sau:
(1+3+5+....+2009) --- ( 2+4+6+.....+2010)
= {{ ((2009 -1)/2 +1) x (2009 +1) } / 2 }} --- {{ (( 2010 - 2) /2+1) x (2010+2)) / 2 }}
= 1010025 --- 1011030
= -1005
giá trị rút gọn là sao pạn?mình ko pit mình chỉ pit kết quả:
C=1-2+3-4+....+2007-2008+2009-2010
=(1-2)+(3-4)+...+(2007-2008)+(2009-2010)
=-1+-1+....+-1+-1
=-1.(-2010-1+1):2
=1.(-2010):2
=1.(-1005)
=-1005
NHân 2B rùi lấy 2B-B ra B
Gọi C =1 + 2 + 2^2 + ...+2^2008
2C =2.(1 + 2 + 2^2 +...+2^2008)
=2 + 2^2 +...+2^2009
2C-C = ( 2 + 2^2 +...+2^2009) - (1 +2 +2^2 +...+2^2008)
= 2^2009 - 1
Ta có :2^2009 -1 /1-2^2009 = -(1-2^2009) / 1-2^2009 = - 1
Vậy B = - 1