tính tổng dãy số sau
C= \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=\(1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
=\(\frac{3^6}{3^6}+\frac{3^5}{3^6}+\frac{3^4}{3^6}+\frac{3^3}{3^6}+\frac{3^2}{3^6}+\frac{3^1}{3^6}+\frac{3^0}{3^6}\)
=\(\frac{3^6+3^5+3^4+3^3+3^2+3+1}{3^6}\)
=\(\frac{729+243+81+27+9+3}{729}\)
=\(\frac{1093}{729}\)
nha.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt biểu thức đó là X
ta có :
\(3X=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3X-X=1-\frac{1}{729}\)
\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi tong trên là A
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)
\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)
\(2A=1-\frac{1}{2187}\)
\(2A=\frac{2186}{2187}\)
\(A=\frac{2186}{2187}:2\)
\(A=\frac{1093}{2187}\)
Vậy tổng A = \(\frac{1093}{2187}\)
\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)
\(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)
=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)
<=> 2y = 3- 1/2187
=> y = \(\frac{3-\frac{1}{2187}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{729}{729}+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{729+243+81+27+9+3+1}{729}\)
\(=\frac{1093}{729}\)
gọi biểu thức trên là A
ta có : A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) (1)
\(\frac{1}{3}\)x A =\(\frac{1}{3}\)+\(\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\) (2)
lấy (1) - (2)
\(\frac{2}{3}xA\)= 1 - \(\frac{1}{2187}\)
\(\frac{2}{3}xA\)= \(\frac{2186}{2187}\)
A = \(\frac{2186}{2187}:\frac{2}{3}\)
A = \(\frac{1093}{729}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{3^n}\left(n\in N\right)\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^n}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^n}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^n}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^n}}{2}\)
Giả sử ABCD là một hình vuông có cạnh là 1 đơn vị. Diện tích hình đó là 1.
Diện tích hình chữ nhật S1 bằng \(\frac{1}{3}\) hình vuông nên có diện tích là:
S1 = \(\frac{1}{3}\)
Chia ba phần còn lại của hình vuông ABCD, ta được hình vuông S2. Diện tích hình S2 bằng\(\frac{1}{9}\)hình vuông ABCD nên:
S2 = \(\frac{1}{9}\)
Tiếp tục chia ba phần con lại của của hình vuông ABCD, ta được hình chữ nhật S3 có diện tích:
S3 = \(\frac{1}{27}\)
Tiếp tục làm như thế và cộng lại, ta có:
S1 + S2 + S3 + S4 + S5 + S6 + ... = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+...\)
Như vậy càng kéo dài tổng diện tích của các hình đó thì tổng ấy sẽ tiến dần đến diện tích hinh vuông ABCD, hay nói cách khác:
S1 + S2 + S3 + S4 + S5 + S6 + ... = SABCD
hoặc \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+...\)= 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{531441}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\right)\)
\(2A=1-\frac{1}{3^{12}}\)
\(2A=\frac{531440}{531441}\)
\(A=\frac{531440}{531441}\div2\)
\(A=\frac{265720}{531441}\)
Chúc bạn học tốt!!!!!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
\(3A=3+1+...+\frac{1}{3^4}\)
\(3A-A=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2A=3-\frac{1}{3^5}\)
\(A=\frac{3-\frac{1}{3^5}}{2}\)
Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)
\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)
\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(S\times2=3-\frac{1}{243}\)
\(S\times2=\frac{728}{243}\)
\(S=\frac{728}{243}\div2\)
\(S=\frac{364}{243}\)
Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))
2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)
A = \(\frac{728}{729}:2=\frac{364}{729}\)
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
= \(\frac{81}{243}+\frac{27}{243}+\frac{9}{243}+\frac{3}{243}+\frac{1}{243}\)
= \(\frac{81+27+9+3+1}{243}\)
= \(\frac{121}{243}\)
Câu 1:
C=1/3+1/9+1/27+1/81+1/243
3×C=3×(1/3+1/9+1/27+1/81+1/243)
3×C=1+1/3+1/9+1/27+1/81
3×C−C=(1+1/3+1/9+1/27+1/81)−(1/3+1/9+1/27+1/81+1/243)
2×C=1−1/243
2×C=242/243
C=242/243:2
C=121/243