`@` `\text {Ans}`

`\downarrow`

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}?\)

Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

`3A=`\(3\times\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

`3A =`\(3+\dfrac{3}{3}+\dfrac{3}{9}+\dfrac{3}{27}+\dfrac{3}{81}+\dfrac{3}{243}+\dfrac{3}{729}\)

`3A =`\(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

`3A - A=`\(\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

`2A =`\(3-\dfrac{1}{729}\)

`2A=`\(\dfrac{2186}{729}\)

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

chiu thua

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{13}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{40}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{121}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{364}{243}+\dfrac{1}{720}\)
\(=\dfrac{1092}{720}=\dfrac{91}{60}\)
hoặc có thể viết dưới dạng thập phân gần bằng \(1,5167\)

Khiêm Nguyễn Gia

Xem lại bài?

\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
Để tính tổng của dãy hình học hữu hạn như trên, chúng ta có thể sử dụng công thức sau:
Tổng \(=a\cdot\dfrac{1-r^n}{1-r}\)
Trong đó:
+ \(a\) là số hạng đầu tiên của dãy (trong trường hợp này, \(a=1\)).
+ \(r\) là hệ số nhân của dãy. Rõ ràng \(r=\dfrac{1}{3}\)
+ n là số lượng số hạng trong dãy (trong trường hợp này, \(n=7\)).
Áp dụng công thức trên cho dãy trên, ta có:
Tổng \(=1\cdot\dfrac{1-\left(\dfrac{1}{3}\right)^7}{1-\dfrac{1}{3}}\)
\(=\dfrac{1-\dfrac{1}{2187}}{\dfrac{2}{3}}=\dfrac{\dfrac{2186}{2187}}{\dfrac{2}{3}}=\dfrac{2186}{2187}\cdot\dfrac{3}{2}=\dfrac{6558}{4374}=\dfrac{1093}{729}\)

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{143}+\dfrac{1}{729}\) (sửa \(\dfrac{1}{720}\) thành \(\dfrac{1}{729}\))

\(A=1+\dfrac{1}{3}+\dfrac{1}{3.3}+\dfrac{1}{3.3.3}+\dfrac{1}{3.3.3.3}+\dfrac{1}{3.3.3.3.3}+\dfrac{1}{3.3.3.3.3.3}\)

\(\Rightarrow\dfrac{1}{3}xA=\dfrac{1}{3}+\dfrac{1}{3.3}+\dfrac{1}{3.3.3}+\dfrac{1}{3.3.3.3}+\dfrac{1}{3.3.3.3.3}+\dfrac{1}{3.3.3.3.3.3}+\dfrac{1}{3.3.3.3.3.3.3}\)

Ta lấy \(A-\dfrac{1}{3}xA\) 

\(=1+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3.3}-\dfrac{1}{3.3}\right)+\left(\dfrac{1}{3.3.3}-\dfrac{1}{3.3.3}\right)+\left(\dfrac{1}{3.3.3.3}-\dfrac{1}{3.3.3.3}\right)+\left(\dfrac{1}{3.3.3.3.3}-\dfrac{1}{3.3.3.3.3}\right)+\left(\dfrac{1}{3.3.3.3.3.3}-\dfrac{1}{3.3.3.3.3.3}\right)-\dfrac{1}{3.3.3.3.3.3.3}\)

\(A-\dfrac{1}{3}xA=1-\dfrac{1}{3.3.3.3.3.3.3}\)

\(Ax\left(1-\dfrac{1}{3}\right)=1-\dfrac{1}{3.3.3.3.3.3.3}\)

\(Ax\dfrac{2}{3}=1-\dfrac{1}{3.3.3.3.3.3.3}\)

\(A=\left(\dfrac{3.3.3.3.3.3.3-1}{3.3.3.3.3.3.3}\right):\dfrac{2}{3}=\left(\dfrac{3.3.3.3.3.3.3-1}{3.3.3.3.3.3.3}\right).\dfrac{3}{2}\)

\(A=\left(\dfrac{2187-1}{2187}\right).\dfrac{3}{2}=\dfrac{2186.3}{2187.2}=\dfrac{1093}{729}\)

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