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 \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)

khó vậy nhỉ

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

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

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

\(2\times A=1-\frac{1}{729}=\frac{728}{729}\)

\(A=\frac{364}{729}\)

1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy: 
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243

tick đúng nha

=364/243

đà đà na đa

= 27/81 + 9/81 + 3/81 + 1/81

= ( 27/81 + 3/81 ) + ( 9/81 + 1/81 )

= 30/81 + 10/81

= 40/81

# Math is easy

Đunga 

Rùi

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

\(A=1-\frac{1}{64}\)

\(A=\frac{63}{64}\)

\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(3B-B=1-\frac{1}{243}\)

\(2B=\frac{242}{243}\)

\(B=\frac{242}{243}\div2\)

\(B=\frac{121}{243}\)

a.A=1/2+1/4+1/8+1/16+1/32+1/64

 A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}\)

= 1 - 1/8 = 7/8

b.B=1/3+1/9+1/27+1/81+1/243

B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)

= 1 - 1/27 = 26/27

F*** you bich

Số số hạng : ( 49 - 1 ) : 2 +1 = 25 

Tổng : ( 49 + 1 ) x 25 : 2 = 625

Khoảng cách các phần tử là : 3 - 1 = 2

có số phần tử là : ( 49 - 1 ) : 2 + 1 = 25 ( phần tử )

Tổng là : ( 49 + 1 ) x 25 : 2 = 625 

Chúc bn học tốt !

Số số hạng của tổng là : 

( 49 - 1 ) : 2 + 1 = 15 (số hạng)

Tổng :

1 + 3 + 5 + 7 + 9 + 11 + 13 + .... + 47 + 49

= ( 1 + 49 ) x 15 : 2

= 50 x 15 : 2

=375

S = 1/3+1/9+1/27+1/81+1/243+1/729+1/2187 ( 1 ) 
Nhân S với 3. Ta có: 
S x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 ( 2 ) 
Trừ ( 2 ) với ( 1 ) ta có: 
S x 3 - S = 1 - 1/ 2187 
2S = 2186/ 2187 
S = 2186/ 2187 : 2 
S = 1093/ 2187 

\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)

= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)

= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)

= \(1+1+1+1+1\)

= \(1\times5\)

= \(5\)

Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :

A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10

A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)

A = 1 + 1 + 1 + 1 + 1

A = 5

Ta có \(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)

\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{3}{8}+\frac{5}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\)

\(=1+1+1+1+1\)

\(=5\)