Tìm giá trị của biểu thức:
1+1/3+1/9+1/27+...+1/729x3
/ là phân số.
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\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(=\frac{2187}{2187}+\frac{729}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)
\(=\frac{2187+729+81+27+9+3+1}{2187}\)
\(=\frac{3037}{2187}\)
Đúng 100%
Lời giải:
Đặt $A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}$
$3\times A=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}$
$3\times A-A=3-\frac{1}{2187}$
$2\times A=3-\frac{1}{2187}=\frac{6560}{2187}$
$A=\frac{6560}{2187}:2=\frac{3280}{2187}$
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729.3}\)
\(A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)
=> \(A=\frac{3^8-1}{2.3^7}\)
Dấu "." là dấu nhân bạn nhé.
Ta có:
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)
\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)
\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)
\(\Rightarrow2A=3-\dfrac{1}{729.3}\)
\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729\cdot3}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3A-A=\left(4+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(2A=3-\frac{1}{3^7}\)
\(A=\frac{ 1}{2}\left(3-\frac{1}{3^7}\right)\)
Gia trị biểu thức là : (3-1/3^7) / 2
Tk mk nha