1+2+4+8+16+32+...+32768
tính cho mk nha
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Đặt A = 1 + 2 + 4 + 8 + ..... + 1024
=> 2A = 2 + 4 + 8 + ..... + 2048
=> 2A - A = 2048 - 1
=> A = 2047
Đặt A = 1 + 2 + 4 + 8 + ... + 1024
=> 2A = 2 + 4 + 8 + ... + 2048
=> 2A - A = 2048 - 1
=> 2A = 2047
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)
=> A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^8}\)
=> 2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^7}\)
=> 2A-A=\(\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^8}\right)\)
=> A=\(1-\frac{1}{2^8}\)
Tính A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A = ....
Giúp mk nha ! đúng mk sẽ tick cho ^_^ !
A =1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A = 64/128 + 32/128 + 16/128 + 8/128 + 4/128 + 2/128 + 1/128
A = 217/218 tick đúng nha
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(A-\frac{1}{2}A=\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{128}-\frac{1}{128}\right)+\left(\frac{1}{2}-\frac{1}{256}\right)\)
\(A=\left(\frac{1}{2}-\frac{1}{256}\right)\times2=1-\frac{1}{128}=\frac{127}{128}\)
Đề sai nha bạn mình sửa luôn
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{32}{1-x^{32}}=VP\left(đpcm\right)\)
2A=1+1/2+1/4+1/8+1/16+1/32+1/64
2A-A=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2+1/4+1/8+1/16+1/32+1/64+1/128)
A=1-1/128
A=127/128
A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
suy ra: 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A - A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128
A = 1 - 1/128 = 127/128
hok tốt
1/2 + 1/4 + 1/8 +1/16 + 1/32 + 1/64 + 1/128
= 2 . ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 )
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 - 1/128 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 ( Rồi giản ước )
= 1
Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729
Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)
\(A=1+2+4+8+16+32+...+32768\)
\(A=1+2+2^2+2^3+2^4+2^5+...+2^{15}\)
\(A=2^{17}-1\)
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