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Tui cần gấp
Đúng tui k nhưng làm hẳn ra nha :)
Nếu lâu thì làm mẫu 1 vài phần OK :)
a) \(2A=2^1+2^2+2^3+...+2^{2010}\)
=> \(2A-A=2^{2011}-2^0\Leftrightarrow A=2^{2011}-1\)
b) \(3B=3+3^2+3^3+3^4+...+3^{101}\)
=> \(3B-B=3^{101}-1\Leftrightarrow2B=3^{101}-1\Leftrightarrow B=\frac{1}{2}\left(3^{101}-1\right)\)
Tương tự Cx4, Dx5
a)A=2^0+2^1+2^2+....+2^2010
ta lay:2A=2^1+2^2+2^3+...+2^2011
ta lay:2A-A=(2^1+2^2+2^3+...+2^2011)-(2^0+2^1+2^3+...+2^2010)
=2^1+2^2+2^3+...+2^2011-2^0-2^1-2^2-2^3-...-2^2010
=2^2011-2^0=2^2011-1=A
Vay A=2^2011-1
A=1+2+22+......+2100
=>2A=2+2223+......+2100+2101
=>2A-A=(2+22+23+....+2101)-(1+2+22+.....+2100)
=>A=2101-1
B=3+32+...+350
2B=32+33+..+351
2B-B=(32+33+......+351)-(3+32+...+350)
B=351-3
A = 20 + 21 + 22 + ... + 22006
2A = 2 + 22 + 23 +...+ 22006 + 22007
2A - A = ( 2 + 22 + 23 + ... + 22006 + 22007 ) - ( 20 + 21 + 22 +...+ 22006 )
A = 22007 - 1
\(\Rightarrow2A=2^1+2^2+2^3+.......+2^{2006}+2^{2007}\)
\(\Rightarrow2A=2^0+2^1+2^2+.....+2^{2006}+\left(2^{2007}-2^0\right)\)
\(\Rightarrow2A=A+\left(2^{2007}-1\right)\)
\(\Rightarrow A=2^{2007}-1\)
Còn phần B bạn làm tương tự: nhân B với 3
Nhớ k nhé
a. 2A= 2+22+23+......+22007 2A-A= (2+22+23+.........+22007)-(20+21+22+..........+22006) A= 22007-20 A=22007-1 b. 3B= 3+32+33+............+3101 3B-B= (3+32+33+.......+3101)-(1+3+32+33+.........+3100) => 2B=3101-1 => B= (3101-1)/2
a. Ta có:
\(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.71\)
\(72^{44}-72^{43}=72^{43}.\left(72-1\right)=72^{43}.71\)
Vì \(72^{44}.71>72^{43}.71\)
\(\Rightarrow72^{45}-72^{44}>72^{44}-72^{43}\)
\(A = 1 + 2 + 2^2 + 2^3+ ... + 2^{63}\)
\(2A=2+2^2+2^3+...+2^{63}+2^{64}\)
\(2A-A=2+2^2+2^3+...+2^{63}+2^{64}-\left(1+2+2^2+2^3+...+2^{63}\right)\)
\(\Rightarrow A=2^{64}-1\)
\(A=2^0+2^1+2^2+.....+2^{1990}\)
\(2A=2\left(2^0+2^1+2^2+.....+2^{1990}\right)\)
\(2A=2^1+2^2+2^3+.....+2^{1991}\)
\(2A-A=\left(2^1+2^2+2^3+.....+2^{1991}\right)-\left(2^0+2^1+2^2+.....+2^{1990}\right)\)
\(A=2^{1991}-2^0=2^{1991}-1\)
\(B=a^0+a^1+a^2+a^3+.....+a^n\)
\(B.a=a^1+a^2+a^3+a^4+.....+a^{n+1}\)
\(B.a-B=\left(a^1+a^2+a^3+a^4+......+a^{n+1}\right)-\left(a^0+a^1+a^2+a^3+.....+a^n\right)\)
\(B.a=a^{n+1}-1\Leftrightarrow B=\dfrac{a^{n+1}-1}{a}\)
\(C=1+3+3^2+.....+3^{50}\)
\(3C=3\left(1+3+3^2+.....+3^{50}\right)\)
\(3C=3+3^2+3^3+.....+3^{51}\)
\(3C-C=\left(3+3^2+3^3+.....+3^{51}\right)-\left(1+3+3^2+.....+3^{50}\right)\)
\(2C=3^{51}-1\Rightarrow C=\dfrac{3^{51}-1}{2}\)