tính A = 2 + 4 + 8 + .......+ 128 + 256
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Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(A=1-\dfrac{1}{128}=\dfrac{127}{128}\)
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a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
a) 2 x 3 x 4 x 8 x 50 x 25 x 125
= (2 x 50) x (4 x 25) x (8 x 125) x 3
= 100 x 100 x 1000 x 3
= 30 000 000
b) 45 x 128 90 x 64
= (45 x 64) x 128 90
= 2880 x 128 90
= 37 123 200
_HT_
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
1/2 + 1/4 + 1/8 + … + 1/128
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + … + 1/64 - 1/128
= 1 - 1/128
= 128/128 - 1/128
= 127/128
Chúc bạn học tốt.
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\(A=2+4+8+...+128+256\)
\(=2+2^2+2^3+...+2^7+2^8\)
=>\(2A=2^2+2^3+2^4+...+2^8+2^9\)
=>\(2A-A=2^2+2^3+...+2^9-2-2^2-...-2^8\)
=>\(A=2^9-2=512-2=510\)
A = 2 + 4 + 8 + ... + 128 + 256
A x 2 = 2 x (2 + 4 + 8 + ... + 128 + 256)
A x 2 = 4 + 8 + 16 + ... + 256 + 512
A x 2 - A = (4 + 8 + 16 + ... + 256 + 512) - (2 + 4 + 8 + ... + 128 + 256)
A = 512 - 2
A = 510
Vậy A = 510