Tính A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
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10 tháng 4 2019
\(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{10}}\)
\(2A=2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)
\(2A+A=\left(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\right)+\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^{10}}\right)\)
\(3A=2-\frac{1}{2^{10}}\)
\(3A=\frac{2^{11}-1}{2^{10}}\)
\(A=\frac{2^{11}-1}{2^{10}.3}\)
11 tháng 6 2016
Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10
A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ... + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)
A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]
A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2)
A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2
A = 1/2 × 5 = 5/2
15 tháng 6 2016
Đặt A = 1/2 - 1/3 - 2/3 + 1/4 + 2/4 + 3/4 - 1/5 - 2/5 - 3/5 - 4/5 + ... + 1/10 + ...+ 9/10
A = 1/2 - ( 1/3 + 2/3) + (1/4 + 2/4 + 3/4) - ( 1/5 + 2/5 + 3/5 + 4/5) + ( 1/6 + 2/6 + ... + 5/6) - ( 1/7 + 2/7 + ... + 6/7) + ( 1/8 + 2/8 + ... + 7/8) - ( 1/9 + 2/9 + ... + 8/9)
A = 1/2 - 1 + [( 1/4 + 3/4) + 2/4] - [(1/5 + 4/5) + (2/5 + 3/5)] + [(1/6+5/6) + ( 2/6 + 4/6) + 3/6] - [(1/7 + 6/7) + (2/7 + 5/7) + (3/7 + 4/7)] + [(1/8 + 7/8) + (2/8 + 6/8) + (3/8 + 5/8) + 4/8)] - [(1/9 + 8/9) + (2/9 + 7/9) + (3/9 + 6/9) + (4/9 + 5/9)] + [(1/10 + 9/10) + ( 2/10 + 8/10) + ( 3/10 + 7/10) + ( 4/10 + 6/10) + 5/10]
A = 1/2 - 1 + ( 1 + 1/2) - 2 + ( 2 + 1/2) - 3 + ( 3 + 1/2) - 4 + ( 4 + 1/2)
A = 1/2 + 1/2 + 1/2 + 1/2 + 1/2
A = 1/2 × 5 = 5/2
DP
14 tháng 8 2017
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
DP
14 tháng 8 2017
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\)
\(A=\left(a-\frac{1}{2}A\right).2=\left(\frac{1}{2}-\frac{1}{2^{11}}\right).2=\frac{1023}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^9}\)
\(2A-A=1-\frac{1}{2^{10}}\)