S=1/2+1/4+1/8+1/16+...+1/128+1/256 = ?
K
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L
7
11 tháng 3 2019
1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1 – 1/2 + 1/2- 1/4 + 1/4 – 1/8 + 1/8 – 1/16 + 1/16 – 1/32 + 1/32 – 1/64 + 1/64 – 1/128 + 1/128 – 1/256 – 1/256 – 1/512
= 1 – 1/512
= 511/512
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
9 tháng 5 2023
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
NT
3
7 tháng 6 2018
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^2}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{1}{2^7}\)
7 tháng 6 2018
E= 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256
2E = 2 ( 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256 )
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=> E = 2E - E
= (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 + 1/256 )
= 1 - 1/256
= 255/256
k nhá, thanks
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 3 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
2A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2A - A = 1 - \(\dfrac{1}{256}\)
A = \(\dfrac{255}{256}\)
PT
1
H9
HT.Phong (9A5)
CTVHS
21 tháng 9 2023
\(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
\(2\times E=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)
\(2\times E-E=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)\)
\(E=1-\dfrac{1}{256}\)
\(E=\dfrac{256}{256}-\dfrac{1}{256}\)
\(E=\dfrac{255}{256}\)
11 tháng 9 2019
Tính \(S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
Dùng sai phân như sau
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)=1-\frac{1}{256}\)
Vậy \(S=1-\frac{1}{256}\)
NA
5
28 tháng 6 2016
\(\text{Đặt }\)\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(=>A=\frac{255}{256}\)
S= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2S= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=>S = 2S-S =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 - 1/256
=1-1/256
=255/256
255/256
ht
và
NHỚ K CHO MIK NHA!!!!!!!!!!!!!!!!!