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Ta gọi tử của phân số B là A ta có:
A=1+2+2^2+2^3+...+2^2008
2A=2 + 2^2 + 2^3 + 2^4 +... + 2^2009
=>A=2^2009 - 1
A=-1 + 2^2009
ta thấy tử là số đối của mẫu =>B=\(\dfrac{-1}{1}\)
Đặt A=1+2+22+...+220081+2+22+...+22008
=>2A=2.(1+2+22+...+220081+2+22+...+22008)
=>2A=2+22+23+...+220092+22+23+...+22009
=>2A-A=(2+22+23+...+220092+22+23+...+22009)-(1+2+22+...+220081+2+22+...+22008)
=>A=22009−122009−1
=>A=(-1).(−2)2009(−2)2009+(-1).1
=>A=(-1).[(−2)2009+1][(−2)2009+1]
=>A=(-1).(1−22009)(1−22009)
=>1+2+22+...+220081+2+22+...+22008/1-2200922009
=(−1).(1−22009)1−22009(−1).(1−22009)1−22009=-1
Giải:
Đặt A=1+2+22+23+...+22008
2A=2+22+23+24+...+22009
2A-A=(1+2+22+23+...+22008)-(2+22+23+24+...+22009)
A =1-22009
Vậy B=1-22009/1-22009=1
Chúc bạn học tốt!
A = 1 + 2 + 2 2 + . . . + 2 2007
2 A = 2 + 2 2 + . . . + 2 2007 + 2 2008
A = 2A - A = ( 2 + 2 2 + . . . + 2 2007 + 2 2008 ) - ( 1 + 2 + 2 2 + . . . + 2 2007 ) = 2 2008 - 1
Vậy A = 2 2008 - 1
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)
\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)
Vậy \(A=\dfrac{2^{2021}-1}{2^{2021}}\)
Ta có : 1/21 + 1/22 + 1/23 + ... + 1/35<1/20+1/20+...+1/20(35 thừa số 1/20)
=> 1/21 + 1/22 + 1/23 + ... + 1/35<1/20.35=35/20=7/4<1/2
=> ĐPCM
Vậy.................
\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
2B = 2 + 22 + 23 + .... + 22009
2B - B = [2 + 22 + 23 + .... + 22009] - [1 + 2 + 22 + ... + 22008]
B = 22009 - 1
Vậy dãy trên bằng:
\(\frac{2^{2009}-1}{1-2^{2009}}=-1\)