Question

a = (1+3+5+7+...+2009+2011)+2012

Rapid va rog ca nu e greu

Answer 1

Obs. Se aplică formula: 

        1+ 3+ 5+ 7+ ...+ ( 2·n+ 1) = ( n+ 1)²


A = ( 1+ 3+ 5+ 7+ ... + 2 009 + 2 011)+ 2 012

A = [1+ (2+1)+ ( 4+ 1)+ ( 6+ 1) + ... +( 2 0008+ 1) + ( 2 010+ 1)] + 2 012

A = [1+ (2·1+1)+ (2·2+ 1)+ (2·3+ 1) + ... +( 2·1 004+ 1) + ( 2·1 005+ 1)] +
 
      +  2 012

A= ( 1 005 + 1)² + 2 012

A= 1 006² + 2 012

A = 1 006²+ 1 006·2

A= 1 006· 1 006 + 1 006·2

A= 1 006·( 1 006+ 2)

A= 1 006·1 008

A= 1 014 048

Answer 2

   
[tex]\displaystyle\\ A = (1+3+5+7+...+2009+2011)+2012\\ \texttt{Calculam numarul termenilor din paranteza: }\\\\ n = \frac{2011 - 1}{2}+1 = \frac{2010}{2}+1 = 1005+1 = 1006\\\\ 1+3+5+7+...+2009+2011 = \frac{n(2011+1)}{2} =\\\\ = \frac{1006 \times 2012}{2} = {1006 \times{1006 = 1006^2[/tex]

[tex]A = (1+3+5+7+...+2009+2011)+2012 =\\\\ 1006^2 + 2012 = 1006^2 + 2\times 1006 = \\\\ =1006(1006+2) =1006\times 1008 = 1014048 [/tex]