Question

Tính nhanh tổng sau

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

ai kb với mình và làm đúng mình tk cho

Answer 1

2186/729 nha bn k mk đi

Answer 2

A=1+1/3+1/9+...+1/729

3A=3+1+1/3+....+1/243

3A-A=(3+1+1/3+...+1/243)-(1+1/3+1/9+...+1/729)

2A=3-1/729

A=(3-1/729)/2

k cho mình nha

Answer 3

Nhân biểu thức trên với 3 :

(1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729) x 3

= 3 + 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243

Trừ biểu thức trên cho biểu thúc gốc ta có:

  3 + 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243

_

  1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

= 3 - 1/729 = 2186/729

Đúng đó

Answer 4

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

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

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

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

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

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

\(A=3-\frac{1}{729}\)

\(A=\frac{2187}{729}-\frac{1}{729}\)

\(A=\frac{2186}{729}\)