calculati sumele:
a)1+3+5+....+2011
b)2005+2+4+6+4008
c)1996+2+4+6+....+3990
d)22+24+26+...+498

a ) 1 + 3 + 5 + ........ + 2011 =

→ aflam cati termeni are suma:

( 2011 - 1 ) : 2 ( ratia) + 1 = 2010 : 2 + 1 = 1005 + 1 = 1 006 termeni are suma

= 1 006 × ( 1 + 2 011 ) / 2 =

= 1 006 × 2012 / 2 =

= 1 006 × 1006 =

= 1 012 036

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b)

2005 + 2 + 4 + 6 + 4008 = 6 025

= 2005 + ( 4 + 6 ) + ( 2 + 4008 ) = → folosesc asociativitatea adunarii

= 2005 + 10 + 4 010 =

= 6025

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c)

1 996 + 2 + 4 + 6 + ......+ 3 990 =

= 1 996 + 2 × ( 1 + 2 + 3 + .........+ 1 995 ) =

= 1 996 + 2 × 1 995 × 1996 / 2 =

= 1 996 + 1 995 × 1 996 =

= 1 996 ( 1 + 1 995 ) =

= 1 996 × 1 996 =

= 3 984 016

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d)

22 + 24 + 26 + ......... + 498 =

= 2 × ( 11 + 12 + 13 + ...... + 249 ) =

→ aflam cati termeni are suma :

249 - 11 + 1= 238 + 1 = 239 de termeni are suma

= 2 × 239 × ( 11 + 249 ) / 2 =

= 239 × 260 =

= 62 140