giải bài toán sau
tính
S=1-1/2+1/4-1/8+1/16-...+1/1024
K
Khách
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Những câu hỏi liên quan
AC
29 tháng 3 2016
Đặt tổng trên là A . Ta có:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
LH
25 tháng 5 2016
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 =1023/1024
vậy A=1023/1024
25 tháng 5 2016
Đặt A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{1024}\) (1)
Ta có: 2A=\(2+1+\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{512}\) (2)
Từ (1) và (2) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)
\(\Rightarrow A=2-\frac{1}{1024}\)
\(\Rightarrow A=\frac{2047}{1024}\)
P
Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
2
31 tháng 12 2023
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
AH
Akai Haruma
Giáo viên
30 tháng 12 2023
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
T
24 tháng 10 2023
Đặt $A=\dfrac12+\dfrac14+\dfrac18+\dfrac{1}{16}+...+\dfrac{1}{1024}$
$A=\dfrac12+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}$
$\dfrac12\cdot A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}}$
$A-\dfrac{1}{2}A=(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}})-(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}})$
$\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{11}}$
$\dfrac{1}{2}A=\dfrac{1}{2}\cdot(1-\dfrac{1}{2^{10}})$
$\Rightarrow A=1-\dfrac{1}{2^{10}}$
Vậy: ...
$Toru$
KP
2
18 tháng 9 2016
\(-1-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)=-1-\frac{1}{1024}=\frac{-1025}{1024}\)
S
4 tháng 3 2016
A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512 + 1/1024
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512
A x 2 - A = 1 + 1/2 - 1/2+ 1/4 -1/4 + 1/8 -1/8 + 1/16 -1/16 + ... + 1/512 - 1/512 - 1/1024
A = 1 - 1/1024
A = 1023/1024
TN
0
NN
1
Ta có :
\(S=1-\frac{1}{2}+\frac{1}{4}-...+\frac{1}{1024}\)
\(S=1-\frac{1}{2}+\frac{1}{2^2}-...+\frac{1}{2^{10}}\)
\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...+\frac{1}{2^{11}}\)
\(S+\frac{1}{2}S=\left(1-\frac{1}{2}+\frac{1}{2^2}-...+\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...+\frac{1}{2^{11}}\right)\)
\(\frac{3}{2}S=1-\frac{1}{2^{11}}\)
\(S=\frac{1-\frac{1}{2^{11}}}{\frac{3}{2}}\)
\(S=\frac{2-\frac{1}{2^{10}}}{3}\)
\(S=\frac{\frac{2^{11}-1}{2^{10}}}{3}\)
Vậy \(S=\frac{\frac{2^{11}-1}{2^{10}}}{3}\)
Chúc bạn học tốt ~
Ta có:
2S = 2.(1-1/2+1/4-1/8+1/16-...+1/1024)
2S = 2/2-1+1/2-1/4+1/8-...+1/512
2S+S = ( 2/2-1+1/2-1/4+1/8-...+1/512)+(1-1/2+1/4-1/8+1/16-...+1/1024)
3S = 2 + 1/1024
3S = 2048/1024+1/1024
3S = 2049/1024
S = 2049/1024 : 3
S = 2049/1024.1/3
S = 683/1024