Tính P = (1/2 + 1/3 + 1/4 + ... + 1/2012) / (2011/1 + 2010/2 + 2009/3 + ... + 1/2011)

Tính P = (1/2 + 1/3 + 1/4 + ... + 1/2012) / (2011/1 + 2010/2 + 2009/3 + ... + 1/2011)

Đề bài: Tính $\text{P = }\frac{\frac{\text{1}}{\text{2}}\text{ + }\frac{\text{1}}{\text{3}}\text{ + }\frac{\text{1}}{\text{4}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2012}}}{\frac{\text{2011}}{\text{1}}\text{ + }\frac{\text{2010}}{\text{2}}\text{ + }\frac{\text{2009}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2011}}}$

Lời giải:

Đặt $B=\frac{\text{2011}}{\text{1}}\text{ + }\frac{\text{2010}}{\text{2}}\text{ + }\frac{\text{2009}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2011}}$

Cộng 1 vào ta được

$B=\left(\frac{\text{2012}}{\text{1}}\text{ + }\frac{\text{2012}}{\text{2}}\text{ + }\frac{\text{2012}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{2012}{\text{2011}}+\frac{2012}{2012}\right)-2012$

$\Rightarrow \text{B = 2012}\left(\frac{\text{1}}{\text{2}}\text{ + }\frac{\text{1}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2011}}\text{ + }\frac{\text{1}}{\text{2012}}\right)\text{ - 2012 + }\frac{\text{2012}}{\text{1}}$

Thay vào P ta được

$P=\frac{\frac{\text{1}}{\text{2}}\text{ + }\frac{\text{1}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2012}}}{2012\left(\frac{\text{1}}{\text{2}}\text{ + }\frac{\text{1}}{\text{3}}\text{ + }\mathrm{...}\text{ + }\frac{\text{1}}{\text{2012}}\right)}$

$\Rightarrow P=\frac{1}{2012}$

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