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a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
~ Chúc bn học tốt ~
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
a) Có A=\(1+3+3^2+3^3+....+3^{100}\)
\(\Rightarrow\)3A =\(3\left(1+3+3^2+3^3+...+3^{100}\right)\)=\(3+3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow2A=3+3^2+3^3+....+3^{101}-1-3-3^2-3^3-....-3^{100}=3^{101}-1\)\(\Rightarrow A=\dfrac{3^{101}-1}{2}\)
Bài b/c/d : bn cứ lm tương tự.
a) A = -1 - 2 - 3 - 4 - ... - 100
A = - ( 1 + 2 + 3 + 4 + ... + 100 )
A = -5050 ( tính dãy số cách đều )
b) B = 1-2 + 3-4 + 5-6 + ... -100
B = ( 1-2 ) + ( 3-4 ) + ( 5-6 ) + ... + ( 99 - 100 )
B = ( -1 ) + ( -1 ) + ( -1 ) + ... + ( -1 )
B = ( -1 ) . 50
B = -50
a)A=1+2+22+...+2100
=>2A=2+22+23+...2101
=>2A-A=(2+22+23+...+2101)-(1+2+22+...+2100)
=>A=2101-1
b)B=3+32+33+...+3100
=>3B=32+33+...+3101
=>3B-B=(32+33+...+3101)-(3+32+...3100)
=>2B-B=3101-3
=>B=(3101-3):2
c)C=1+2+4+8+16+...+8192
=>C=1+2+22+23+...213
=>2C=2+22+23+...+214
=>2C-C=(2+22+...+214)-(2+22+...+213)
=>C=214-2
d)D=4+42+43+...+4n
=>4D=42+43+...+4n+1
=>4D-D=(42+43+...+4n+1)-(4+42+...+4n)
=>3D=4n+1-4
=>D=(4n+1-4):3
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
a, A = 4 + 4^2 + 4^3 + ... + 4^n
=> 4A = 4.(4 + 4^2 + 4^3 + ... + 4^n)
=> 4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1
=> 3A = 4A - A = (4^2 + 4^3 + 4^4 + ... + 4^n+1) - ( 4 + 4^2 + 4^3 + ... + 4^n)
=> 3A = 4^n+1 - 4
=> A = \(\frac{4^{n+1}-4}{3}\)
Vậy A = ..................
b, B = 1 + 3 + 3^2 + ... + 3^100
=> 3B = 3.(1 + 3 + 3^2 + ... + 3^100)
=> 3B = 3 + 3^2 + 3^3 + ... + 3^101
=> 2B = 3B - B =(3 + 3^2 + 3^3 + ... + 3^101) -(1 + 3 + 3^2 + ... + 3^100)
=> 2B = 3^101 - 1
=> B = \(\frac{3^{101}-1}{2}\)
Vậy B = ......................
A = 4 + 4^2 + 4^3 + ... + 4^n
4A = 4^2 + 4^3 + 4^4 + ... + 4^n+1
4A - A = ( 4^2 + 4^3 + 4^4 + ... + 4^n+1 ) - ( 4 + 4^2 + 4^3 +...+4^n)
3A = 4^n+1 - 4
A = 4^n+1 - 4/3