\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(=\frac{2187}{2187}+\frac{729}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)

\(=\frac{2187+729+81+27+9+3+1}{2187}\)

\(=\frac{3037}{2187}\)

Đúng 100%

toán lớp 4 chưa học bình phương -.-

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:

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)

\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)

\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)

\(\Rightarrow2A=3-\dfrac{1}{729.3}\)

\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)

Đáp án A

Đáp án : A

Đáp án C