2 replies to this topic

[lovemath99]

Tính $A=3-3^2+3^3-3^4+..+3^{2003}3^{2004}$

Edited by lovemath99, 28-01-2015 - 19:52.

[Mary Huynh]

Tính $A=3-3^2+3^3-3^4+..+3^{2003}-3^{2004}$

Dấu $ ''-''$ chứ nhỉ ?

Nếu thế thì : 
$3A=3^{2}-3^{3}+3^{4}-....+3^{2004}-3^{2005}$

$\Rightarrow 4A=3-3^{2005}$

$\Rightarrow A=\frac{3-3^{2005}}{4}$

Edited by Mary Huynh, 30-01-2015 - 12:11.

[phungvip]

A = 3 - 3² + .... + 3²⁰⁰³ -3²⁰⁰⁴

       3A =  3² - .... - 3²⁰⁰³ -3²⁰⁰⁴  - 3²⁰⁰⁵

4A = 3 + 3²⁰⁰⁵

A= (3 +3²⁰⁰⁵) : 4