Question

x=?
x=1+3+5+...+2011
:)

Answer 1

Folosind formula

$1+3+5+...+(2n-1)=n^2$

si observand ca 2n-1=2011⇒n=1006, avem:

$1+3+5+...+2011=1+3+5+...+(2\cdot1006-1)=1006^2$