Question

[Nguyễn Hải Vân]

A=1/2+3/2+(3/2)^2+(3/2)^3+...+(3/20)62012;B=(3/2)^2013 Tính B-A

Answer 1

[Akai Haruma]

Lời giải:
$A=\frac{1}{2}+\frac{3}{2}+(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2012}$

$\frac{3}{2}A=\frac{3}{4}+(\frac{3}{2})^2+(\frac{3}{2})^3+(\frac{3}{2})^4+...+(\frac{3}{2})^{2013}$

$\Rightarrow \frac{3}{2}A-A=(\frac{3}{2})^{2013}+\frac{3}{4}-\frac{1}{2}-\frac{3}{2}$

$\Rightarrow \frac{1}{2}A=(\frac{3}{2})^{2013}-\frac{5}{4}$

$A=2(\frac{3}{2})^{2013}-\frac{5}{2}$

$\Rightarrow B-A=(\frac{3}{2})^{2013}-2(\frac{3}{2})^{2013}+\frac{5}{2}=\frac{5}{2}-(\frac{3}{2})^{2013}$