Tinh: -1 - 1/2 - 1/4 - 1/8 - ... - 1/1024 .
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
S= 1/2+1/2^2+1/2^3+.....+1/2^10
-
1/2x S =1/2^2+1/2^3+....+1/2^10+1/2^11
1/2xS=1/2-1/2^11
S=(1/2-1/2^11) :1/2
=(1/2-1/2^11)x2
= 1-1/2^10 =1-1/1024 =1023/1024
S=\(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> 2S=\(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2S - S = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> S = \(1-\frac{1}{2^{10}}\)
Đáp án này đúng 100% nha
mình làm trước
\(A=\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
Đặt \(B=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
=>2B=1/2+1/4+...+1/512
=>B=1/2-1/1024
A=1/2-1/2+1/2024=1/1024
Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)
nhân cả cụm đó với 2 rồi trừ đi cụm ban đầu thì còn là -2-1/1024 =-2049/1024
A/5+10+15+...+1500
=5+10+15+...+1500 ta có:1500-5:5+1=300(số hạng)
=(5+1500)x300:2=225750
Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
\(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-....-\dfrac{1}{1024}\)
\(A=-1-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
Đặt:
\(B=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2B=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(B=1-\dfrac{1}{2^{10}}\)
Thay vào \(A\)
\(A=-1-\left(1-\dfrac{1}{2^{10}}\right)\)
\(A=-1-1+\dfrac{1}{2^{10}}\)
\(A=-2+\dfrac{1}{2^{10}}\)