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Đặt A= 1/3+1/9+1/27+1/81+1/243
A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5
3A=1+1/3+1/3^2+1/3^3+1/3^4
3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5
2A=1-1/3^5
2A=242/243
A=121/243
\(\text{Đ}\text{ặt}:A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=3-\frac{1}{729}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2=\frac{1093}{729}\)
A=1/3+1/9+...+1/243
=>3A=1+1/3+...+1/81
=>2A=1-1/243=242/243
=>A=121/243
đặt biểu thức đó là X
ta có :
\(3X=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3X-X=1-\frac{1}{729}\)
\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
1+1/3 + 1/9 + 1/27 + 1/81 + 1/243
=243/243+81/243+27/243 +3/243 +1/243
=\(\dfrac{243+81+27+3+1}{243}\)
=355/243
Lời giải:
$A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}$
$3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}$
$3A-A=3-\frac{1}{243}$
$2A=\frac{728}{243}$
$A=\frac{364}{243}$
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)