Question

calculati √1+3+5+...+2009+2011/1+3+5+...+1003+1005 radical din tot

Answer 1

1+3+5+...+2009+2011=1006(termeni)·2012/2=1006²
1+3+5+...+1003+1005=503(termeni)·1006/2=503²
√(1006²)/(503²)=1006/503=2

Answer 2

$1+3+5+...+n = (\frac{n+1 }{2})^{2} \\ \\ 1+3+5+...+2011= (\frac{2011+1 }{2})^{2} = (\frac{2012 }{2})^{2} = 1006^{2} \\ \\ 1+3+5+...+1005 = (\frac{1006 }{2})^{2}= 503^{2} \\ \\ \Rightarrow \sqrt{ \frac{ 1+3+5+...+2011}{1+3+5+...+1005} } = \sqrt{ \frac{1006^{2} }{503^{2} } }= \frac{1006}{503} = 2$