Câu hỏi của nguyenthuthao

Question

cho a= 3^2+3^3+3^4+...+3^2005+3^2006 .

tìm x để:2A+3=3x

Nguyễn Thanh Hằng · Accepted Answer

$A=3^2+3^3+3^4+...........+3^{2006}$

$\Leftrightarrow3A=3^3+3^4+3^5+................+3^{2006}+3^{2007}$

$\Leftrightarrow3A-A=\left(3^3+3^4+3^5+.........+3^{2007}\right)-\left(3^2+3^3+..........+3^{2006}\right)$

$\Leftrightarrow2A=3^{2007}-3^2$

Đến đây thì tịt ngẩm!Câu trả lời: $A=3^2+3^3+3^4+...........+3^{2006}$

$\Leftrightarrow3A=3^3+3^4+3^5+................+3^{2006}+3^{2007}$

$\Leftrightarrow3A-A=\left(3^3+3^4+3^5+.........+3^{2007}\right)-\left(3^2+3^3+..........+3^{2006}\right)$

$\Leftrightarrow2A=3^{2007}-3^2$

Đến đây thì tịt ngẩm!

Nguyễn Huy Tú · Answer

$A=3^2+3^3+...+3^{2006}$

$\Rightarrow3A=3^3+3^4+...+3^{2007}$

$\Rightarrow3A-A=\left(3^3+3^4+...+3^{2007}\right)-\left(3^2+3^3+...+3^{2006}\right)$

$\Rightarrow2A=3^{2007}-3^2$

Do $2A+3=3x$

$\Rightarrow3^{2007}-3^2+3=3x$

$\Rightarrow3^{2007}-6=3x$

$\Rightarrow x=3^{2006}-2$

Vậy...Câu trả lời: $A=3^2+3^3+...+3^{2006}$