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

=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)

=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)  ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))

2M = 1 - \(\frac{1}{6561}\)

2M = \(\frac{6560}{6561}\)

=> M = \(\frac{3280}{6561}\)

\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)

\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)

\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)

\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)

\(\Rightarrow2M=1-\frac{1}{3^8}\)

\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)

Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)

\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)

\(\Rightarrow3M=3\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\right)\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{6561}\right)\)

\(\Rightarrow2M=1-\frac{1}{6561}\)

\(\Rightarrow2M=\frac{6560}{6561}\)

\(\Rightarrow M=\frac{3280}{6561}\)

3M=1+1/3+1/9+...+1/2187

2M=3M-M

2M=1-1/6561

2M=6560/6561

M=3280/6561

ta có :

= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )

= 59050 + 2190 + 6570 + 270 + 810

= 59050 + ( 2190 + 810 ) + 6570 + 270

= 59050 + 3000 + 6570 + 270

= 59050 + ( 3000 + 6570 ) + 270

= 59050 + 9570 + 270

= 68620 + 270

= 68890

68890

Kết quả là 68890

Nhớ trả lời cho mình

Ta có: 

\(S=3.2^0-3^1+3.2^1-3^2+3.2^2+3.2^3-3^3+3.2^4-3^4+...-3^7+3.2^{10}+3.2^{11}-3^8+3.2^{12}\)

\(=3.\left(2^0+2^1+2^2+2^3+2^4+...+2^{10}+2^{11}+2^{12}\right)-\left(3^1+3^2+3^3+...+3^7+3^8\right)\)

Đặt: \(A=2^0+2^1+2^2+...+2^{11}+2^{12}\)

=> \(2.A=2^1+2^2+2^3+...+2^{12}+2^{13}\)

=> \(2.A-A=2^{13}-2^0\)

\(\Rightarrow A=2^{13}-1=8191\)

Đặt: \(B=3^1+3^2+3^3+...+3^8\)

 \(\Rightarrow3.B=3^2+3^3+3^4+...+3^9\)

=> \(3B-B=3^9-3^1=19680\)

=> \(2B=19680\Rightarrow B=9840\)

=> S=3.A-B=3.8191-9840=14733

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

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

    \(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{19683}\)

 \(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right)-\left(\frac{1}{9}+\frac{1}{27}+...+\frac{1}{19683}\right)\)

\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{19683}\)

\(A=\frac{4840}{9683}:\frac{2}{3}=\frac{7260}{9683}\)

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)

\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)

\(\Rightarrow3A=3.\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\) \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)

\(\Rightarrow2A=1-\frac{1}{3^8}\) \(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)

k cho mik đi mn!Nguyễn Như Quỳnh!

1+3+9+27+....+2187+6561

đặt A = 1+3+9+27+....+2187+6561 

=>A = 3⁰ + 3¹ + 3² + 3³ + .. . +3⁷ + 3⁸ 

 3A = 3¹ + 3² + 3³  + ... + 3⁸ + 3⁹

3A - A = (3¹ + 3² + 3³  + ... + 3⁸ + 3⁹)-(3⁰ + 3¹ + 3² + 3³ + .. . +3⁷ + 3⁸ )

2A = 3⁹ - 1 

A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)

1+3+9+27+....+2187+6561

Câu hỏi của nguyenphucthang - Toán lớp 4 - Học toán với OnlineMath