Sửa đề: A=1/3+1/9+1/27+...+1/6561

=1/3+1/3^2+1/3^3+...+1/3^8

=>3A=1+1/3+...+1/3^7

=>3A-A=1-1/3^8

=>\(2A=\dfrac{3^8-1}{3^8}\)

=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)

Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)

\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)

\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)

\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)

\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)

\(2S=\dfrac{4132}{6561}\)

\(S=\dfrac{2066}{6561}\)

Đặt Tử số là A ta có

\(2A=2+2^2+2^3+2^4+..+2^{2016}\)

\(A=2A-A=2^{2016}-1\)

\(\Rightarrow S=\frac{2^{2016}-1}{1-2^{2016}}=\frac{-\left(1-2^{2016}\right)}{1-2^{2016}}=-1\)

\(S=\frac{1+2+2^2+2^3+...+2^{2015}}{1-2^{2016}}\)

\(\Rightarrow2S=\frac{2\left(1+2+2^2+2^3+...+2^{2015}\right)}{1-2^{2016}}\)

\(\Rightarrow2S=\frac{2+2^2+2^3+2^4+...+2^{2016}}{1-2^{2016}}\)

\(\Rightarrow2S-S=\frac{2+2^2+2^3+2^4+...+2^{2016}}{1-2^{2016}}-\frac{1+2+2^2+2^3+...+2^{2015}}{1-2^{2016}}\)

\(\Rightarrow S=\frac{2^{2016}-1}{1-2^{2016}}=-1\)

Khi nào có bài khó thì cứ đăng lên nhé, mình sẽ giúp ^.^

Minh nghi la bai Viet Nam nhung chuyen di !

bài "VIỆT NAM ƠI" (việt nam ới việt nam ơi,tự hào hát mãi lên VIỆT NAM ƠI )

Mình nghĩ bạn nên múa bài"Vươn cao Việt Nam"vì bài này 20/11 năm ngoái chúng mình múa bài này được giải nhất.

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)

\(3B=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{6561}\right)\)

\(3B=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)

\(2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{6561}\right)\)

\(2B=0+0+...+1-\dfrac{1}{6561}\)

\(2B=1-\dfrac{1}{6561}\)

\(B=\left(1-\dfrac{1}{6561}\right):2\)

\(B=\dfrac{6560}{6561}:2\)

\(B=\dfrac{3280}{6561}\)

{3280}{6561}

\(\)

hay quá tiểu thư họ nguyễn

Ờ , hi bn đã trở lại .

Mong bn sẽ đóng góp nhiều hơn nữa cho hoc24.vn

xl bạn mình nỡ báo cáo rồi

Giải câu 3 trước nha!

M = 6¹⁰⁰⁴.6¹⁰⁰⁴

= 6^1004+1004

⁼ 6²⁰⁰⁸

Bài làm:

Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)

=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\) 

=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)

=> \(A=\frac{3^8-1}{3^8.2}\)

                          Bài làm :

Ta có :

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

\(\Rightarrow3\times A=\frac{1\times3}{3}+\frac{1\times3}{9}+\frac{1\times3}{27}+...+\frac{1\times3}{6561}\)

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}\)

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\left(\frac{1}{6561}-\frac{1}{6561}\right)\)

\(3\times A=1+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}\right)-\frac{1}{6561}\)

\(3\times A=1+A-\frac{1}{6561}\)

\(\Rightarrow2\times A=1-\frac{1}{6561}\)( Trừ bỏ A ở cả 2 vế )

\(2\times A=\frac{6560}{6561}\)

\(A=\frac{6560}{6561}\div2=\frac{3280}{6561}\)

Vậy A=3280/6561

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\)

\(3\times A=2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{729}+\frac{2}{2187}\)

\(3\times A-A=\left(2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{2187}\right)-\left(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\right)\)

\(2\times A=2-\frac{2}{6561}\)

\(A=\frac{6560}{6561}\)

\(=\frac{1}{2}\frac{2}{3}\frac{3}{4}\frac{4}{5}\frac{5}{6}\frac{6}{7}\frac{7}{8}\frac{8}{9}....\frac{2002}{2003}\frac{2003}{2004}\)

\(=\frac{1}{2004}\)