Răspuns:
Sunt egale.
Explicație pas cu pas:
[tex] \displaystyle a = \frac{1}{2}+\frac{2}{2} +\cdots+\frac{2011}{2}= \frac{1}{2}\sum_{n=1}^{2011}n = \frac{1}{2}\cdot\frac{2011\cdot 2012}{2}= \frac{1006\cdot 2011}{2}=503\cdot 2011\\ \\ b = \Bigg(\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{2012}\Bigg) + \Bigg ( \frac{2}{3}+\frac{2}{4}+\cdots + \frac{2}{2012}\Bigg) + \cdots + \Bigg(\frac{2010}{2011}+\frac{2010}{2012}\Bigg) + \frac{2011}{2012}\\ \\ \text{Voi grupa termenii cu acelasi numitor}\\ \\ b = \Bigg(\frac{1}{2}\Bigg) + \Bigg(\frac{1}{3}+\frac{2}{3}\Bigg)+\Bigg(\frac{1}{4}+\frac{2}{4}+ \frac{3}{4}\Bigg) + \cdots + \Bigg(\frac{1}{2012}+\frac{2}{2012}+\cdots+\frac{2011}{2012}\Bigg)\\ \\ b = \sum_{n=2}^{2012}\sum_{k=1}^{n-1}\frac{k}{n}\\ \\ b = \sum_{n=2}^{2012}\frac{1}{n}\sum_{k=1}^{n-1}k\\ \\ b = \sum_{n=2}^{2012}\frac{1}{\not{n}}\cdot \frac{(n-1)\not{n}}{2} \\ \\ b = \frac{1}{2}\sum_{n=2}^{2012}(n-1) \\ \\ b = \frac{1}{2}\sum_{n=1}^{2011}n \\ \\ b = \frac{1}{2}\cdot \frac{2011\cdot 2012}{2} \implies a = b[/tex]
