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\(2A\)\(=\)\(2^2+2^3+2^4+......+2^{2009}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)\(A=2^{2009}-1\)
\(B-A=2009-\left(2009-1\right)\)
\(B-A=1\left(ĐPCM\right)\)
a giải luôn cho e nhé
7A=7+72+73+...+72008
7A-A=[7+72+73+...+72008]-[1+7+72+..+72007]
6A=72008-1
A=72008-1/6
b,Tương tư nhân B vs 4 là ra
Mình chỉ trả lời được 2 câu đầu thôi nhé:
a.A= \(1+7+7^2+7^3+...+7^{2007}\)
A.7 = \(7+7^2+7^3+7^4+...+7^{2008}\)
A7-A = \(\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
A6 =\(7^{2008}-1\)
\(\Rightarrow A=7^{2008}-1\)
Câu còn lại làm tương tự bạn nhé
A = 1 + 2+22+23+....+22008
2A = 2 + 22+23+24+.....+22009
A = 2A - A = 22009 - 1
Vậy B - A = 22009 - (22009 - 1) = 22009 - 22009 + 1 = 1
A = 2+22+23+....+22008
2A = 22+23+24+.....+22009
A = 2A - A = 22009 - 2
Vậy B - A = 22009 - (22009 - 2) = 22009 - 22009 + 2 = 2
a)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{59}.3\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{58}.7\)
\(=7\left(2+2^4+2^{58}\right)⋮7\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=2.15+2^5.15+...+2^{57}.15\)
\(=15\left(2+2^5+2^{57}\right)⋮15\)
b) \(B=1+5+5^2+5^3+...+5^{96}+5^{97}+5^{98}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)
\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+..+5^{96}\left(1+5+5^2\right)\)
\(=31+5^3.31+...+5^{96}.31\)
\(=31\left(1+5^3+...+5^{96}\right)⋮31\)
\(A=2+2^2+2^3+...+2^{2008}\)
\(2A=2^2+2^3+2^4+...+2^{2009}\)
\(A=2A-A=2^{2009}-1\)
Vậy \(B-A=2^{2009}-\left(2^{2009}-1\right)=2^{2009}-2^{2009}+1=1\)
\(A=2^{2010}-1\) cái này cần trả lời tiếp
\(\left(A+1\right).5^{2010}=\left(2^{2010}-1+1\right).5^{2010}=2^{2010}.5^{2010}=10^{2010}=\left(10^{1005}\right)^2=dpcm\)
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
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
A = 1+2+22+23+...+22008
2A = 2+22+23+24+...+22009
A = 2A - A = 22009 - 1
=> B - A = 22009 - (22009-1) = 22009 - 22009 + 1
=> B - A = 1 (đpcm)
A=1+2+2^2+2^3+...+2^2008