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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

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

Lấy 3A - A ta được :

\(2A=1-\dfrac{1}{6561}=\dfrac{6560}{6561}\Leftrightarrow A=\dfrac{6560}{6561}:2\)

\(\Leftrightarrow A=\dfrac{6560}{6561}.\dfrac{1}{2}=\dfrac{3280}{6561}\)

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}\)

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

\(3A-A=\left[1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right]-\left[\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right]\)

\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)

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

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

Vậy : ...

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

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

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

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

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

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

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

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

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

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

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

{3280}{6561}

\(\)

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

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

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

\(2A=\dfrac{6560}{6561}\)

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

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