Giúp mình bài toán này với
1+1\2+1\4+1\8+...+1\1024
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Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(\Rightarrow A=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(B=2+2^2+2^3+...+2^{10}\)
\(2B=2^2+2^3+...+2^{11}\)
\(2B-B=\left(2^2+2^3+...+2^{11}\right)-\left(2+2^2+...+2^{10}\right)\)
\(B=2^{11}-2\)
a)
Từ 1 đến 23 gồm có (29-1) : 2 + 1 = 15 số
15 số gồm có 7.5 cặp
Mà từng cặp số có tổng bằng 30 (tính từng cặp số ở hai đầu)
Vậy S = 30 * 7.5
= 225
=
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2014}\)
\(\Rightarrow-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow-2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow-2A-\left(-A\right)=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)
\(-A=2-\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2^{10}}-2\)
Đặt: \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{4}{2}A=\dfrac{4}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(3+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(3-1-\dfrac{1}{1024}\right)\)
\(\Rightarrow A=2-\dfrac{1}{1024}\)
\(\Rightarrow A=\dfrac{2047}{1024}\)
1+(1)/(2)+(1)/(4)+(1)/(8)+...+(1)/(512)+(1)/(1024)
A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
A x 2 = 1 - 1/1024 + A
A x 2 - A = 1 - 1/1024
A = 1 - 1/1024
A = 1023 /1024
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
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) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)
\(\Rightarrow A=2-\frac{1}{1024}\)
\(\Rightarrow A=\frac{2047}{1024}\)