Question

"supra" este "/"

Sa se compare numerele:
a=1/2+2/2+3/2+...+2011/2

b=(1/2+1/3+1/4+...+1/2012)+(2/3+2/4+2/5+...+2/2012)+(3/4+3/5+3/6+...+3/2012)+...+(2009/2010+2009/2011+2009/2012)+(2010/2011+2010/2012)+2011/2012

VA ROG MULT!!!!!

Answer 1

Răspuns:

a=b

Explicație pas cu pas:

se aplica repetat suma Gauss

Answer 2

Răspuns:

Sunt egale.

Explicație pas cu pas:

$\displaystyle a = \frac{1}{2}+\frac{2}{2} +\cdots+\frac{2011}{2}= \frac{1}{2}\sum_{n=1}^{2011}n = \frac{1}{2}\cdot\frac{2011\cdot 2012}{2}= \frac{1006\cdot 2011}{2}=503\cdot 2011\\ \\ b = \Bigg(\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{2012}\Bigg) + \Bigg ( \frac{2}{3}+\frac{2}{4}+\cdots + \frac{2}{2012}\Bigg) + \cdots + \Bigg(\frac{2010}{2011}+\frac{2010}{2012}\Bigg) + \frac{2011}{2012}\\ \\ \text{Voi grupa termenii cu acelasi numitor}\\ \\ b = \Bigg(\frac{1}{2}\Bigg) + \Bigg(\frac{1}{3}+\frac{2}{3}\Bigg)+\Bigg(\frac{1}{4}+\frac{2}{4}+ \frac{3}{4}\Bigg) + \cdots + \Bigg(\frac{1}{2012}+\frac{2}{2012}+\cdots+\frac{2011}{2012}\Bigg)\\ \\ b = \sum_{n=2}^{2012}\sum_{k=1}^{n-1}\frac{k}{n}\\ \\ b = \sum_{n=2}^{2012}\frac{1}{n}\sum_{k=1}^{n-1}k\\ \\ b = \sum_{n=2}^{2012}\frac{1}{\not{n}}\cdot \frac{(n-1)\not{n}}{2} \\ \\ b = \frac{1}{2}\sum_{n=2}^{2012}(n-1) \\ \\ b = \frac{1}{2}\sum_{n=1}^{2011}n \\ \\ b = \frac{1}{2}\cdot \frac{2011\cdot 2012}{2} \implies a = b$