tính hợp lí tổng sau 1 + 1/2 + 1/4 + 1/8 + ... +1/1024
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:
2/9+6/27+8/36+12/54+16/72+18/81=
2/9+2/9+2/9+2/9+2/9+2/9=
2/9*6=
12/9=
4/3
Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3
b)
Ta có:
1-2/5=3/5
1-2/7=5/7
1-2/9=7/9
...
1-2/99=97/99
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=
3/5*5/7*7/9*...*97/99=
(3*5*7*...*97)/(5*7*9*...*99)=
3/99=
1/33
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33
c)
Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:
S=1/2+1/4+1/8+1/16+...+1/1024
S*2=1+1/2+1/4+1/8+...+1/512
S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)
S=1-1/1024
S=1023/1024
Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024
![](https://rs.olm.vn/images/avt/0.png?1311)
2A=1+1/2+1/4+1/8+.........+1/512
2A-A=(1+1/2+1/4+1/8+....+1/512)-(1/2+1/4+1/8+.....+1/1024)
A=1-1/1024
A=1023/1024
vậy A=1023/1024
A = (1/2+1/64)+(1/4+1/128)+(1/8+1/256)+(1/16+1/512)+(1/32+1/1024)
= 1/2(1+1/32)+1/4(1+1/32)+1/8(1+1/32)+1/16(1+1/32)+1/32(1+1/32)
= (1+1/32) x (1/2+/14+1/8+1/16+1/32)
= (33/32) x (31/32) = 1023/1024
![](https://rs.olm.vn/images/avt/0.png?1311)
s=1/21 + 1/22+1/23+.........+1/210
2s= 1/22 +1/23+.........+1/210+1/211
2s-s= ( 1/22 +1/23+.........+1/210+1/211 ) -(1/21 + 1/22+1/23+.........+1/210)
s= 1/211 -1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dãy số đó có số số hạng là :
( 1/1024 - 1 ) :
( 1 + 1/1024 ) *
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi tổng cần tính là A.
Ta có:
\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\\ A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\\ 2A=2\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ 2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ A=2-\dfrac{1}{2^{10}}\)