Question

A)1+2+3+.....+2003
B)1+2+3+....+2004

Answer 1

suma lui Gauss

(n fiind ultimul termen al sumei)

a.) S= n(n+1)/2= 2003(2003+1)/2= (2003•2004)/2= 4014012/2= 2007006

b.) S= n(n+1)/2= 2004(2004+1)/2= (2004•2005)/2= 4018020/2= 2009010

Answer 2

Răspuns:

Explicație pas cu pas:

a)   1 + 2 + 3 + ..... + 2003 = 2003 × ( 1 + 2003) : 2 = 2003 × 2004 : 2 = 2003 × 1002 = 2007006

=> aplic formula sumei lui Gauss: n ( n + 1 ) : 2; unde n = numarul termenilor consecutivi ai sumei

b) ( 1 + 2 + 3 + ...... + 2003) + 2004 = 2 007 006 + 2004 = 2009010

sau:

1 + 2 + 3 + ..... + 2004 = 2004 x ( 1 + 2004 ) : 2 = 1002 x 2005 =2 009 010