Tính nhanh:1/3+1/3^2+1/3^3+....+1/3^8
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A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
3A = \(1+3+\frac{1}{3^2}+...+\frac{1}{3^7}\)
2A = 3A - A = \(1-\frac{1}{3^8}\)
=> A = \(\frac{1-\frac{1}{3^8}}{2}\)
Ta có: \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^8}\)
\(\Rightarrow A=\left(1-\frac{1}{3^8}\right)\div2\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^8\times2}\)
A=1/3+1/3^2+1/3^3+...+1/3^8
3A=1+1/3+1/3^2+...+1/3^7
3A-A=1-1/3+1/3-1/3^2+1/3^2-1/3^3+...+1/3^7-1/3^8
2A=1-1/3^8
2A=6560/6561
Suy ra A=3280/6561
nho k cho minh voi nhe
1) \(\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\right)\)
\(=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{3}{17}-\frac{3}{17}\right)-\left(\frac{4}{3}-\frac{1}{15}\right)\)
\(=1+0-\frac{19}{15}\)
\(=-\frac{14}{15}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>3A=\(1+\frac{1}{3}+...+\frac{1}{3^7}\)
=>3A-A=\(\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
=>2A=\(1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=>A=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)
A=1/3+1/32+1/33+...+1/38
=>3A=1+1/3+1/32+...+1/37
=>3A-A=(1+1/3+1/32+...+1/37)-(1/3+1/32+1/33+...+1/38)
=>2A=1+1/3+1/32+...+1/37-1/3-1/32-1/33-...-1/38
=1-1/38
=\(\frac{6550}{6561}\)
=>A=\(\frac{6560}{6561}:2=\frac{6560}{6561}.\frac{1}{2}=\frac{3280}{6561}\)
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561