Tính tổng:
A=(1-1/2)+(1-1/4)+(1-1/8)+...+(1-1/512)+(1+1/1024)
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Dãy số đó có số số hạng là :
( 1/1024 - 1 ) :
( 1 + 1/1024 ) *
Đặt A=1/2+1/4+1/8+..+1/1024
Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)
Ax2-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 biểu thức đã cho = 1023/1024
=1-(1/2+1/2-1/4+1/4-1/8+1/8...-1/1024+1/1024-1/1024)
=1-1/1024
=1023/1024
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1028}\)
\(=1-\frac{1}{1028}\)
\(=\frac{1027}{1028}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
\(A=\frac{2^{10}-1}{2^{10}}\)
Tham khảo nhé~
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}\)
vậy \(A=\frac{2^{10}-1}{2^{10}}\)
A = 1/2 + 1/4 + 1/8 + ... + 1/1024
2A = 1 + 1/2 + 1/4 + ... + 1/512
2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)
A = 1 - 1/1024
A = 1023/1024
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)
\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)