calcul sume...remember?
Pun aici repede, repede , o problema (rezolvata ...sa nu aiba nimeni emotii ;-) de clasa a VII, care mi-a dat de furca, atit mie cit si prietenilor mei , toti inteligenti si cu scoli destepte absolvite intr-un trecut mai mult sau mai putin indepartat.
Sa se compare sumele:
a=1-1/2+1/3-1/4+.......................+1/2004
b=1/1003+1/1004+.....................+1/2004
Rezolvare:
a = (1+1/3+1/5+.........+1/2003) – (1/2+1/4+1/6+................+1/2004)
a = (1+1/3+1/5+.........+1/2003) – 2 (1/2+1/4+1/6+................+1/2004) + (1/2+1/4+1/6+................+1/2004)
a =1+1/3+1/5+.........+1/2003 – 1 -1/2-1/3-1/4-1/5-...........-1/1002 + 1/2+1/4+1/6+................+1/2004
a = 1+1/3+1/5+.........+1/2003 – 1 -1/2-1/3-1/4-1/5-...........-1/1002 + 1/2+1/4+1/6+................+1/2004
Reducind termenii rezulta:
a=1/1003+1/1005+............+1/2003+1/1004+1/1006+..................+1/2004
a=1/1003+1/1004+1/1005+..............+1/2004
=> a=b
Sper sa-i foloseasca vreunui studios cindva, undeva...:-)
Sa se compare sumele:
a=1-1/2+1/3-1/4+.......................+1/2004
b=1/1003+1/1004+.....................+1/2004
Rezolvare:
a = (1+1/3+1/5+.........+1/2003) – (1/2+1/4+1/6+................+1/2004)
a = (1+1/3+1/5+.........+1/2003) – 2 (1/2+1/4+1/6+................+1/2004) + (1/2+1/4+1/6+................+1/2004)
a =1+1/3+1/5+.........+1/2003 – 1 -1/2-1/3-1/4-1/5-...........-1/1002 + 1/2+1/4+1/6+................+1/2004
a = 1+1/3+1/5+.........+1/2003 – 1 -1/2-1/3-1/4-1/5-...........-1/1002 + 1/2+1/4+1/6+................+1/2004
Reducind termenii rezulta:
a=1/1003+1/1005+............+1/2003+1/1004+1/1006+..................+1/2004
a=1/1003+1/1004+1/1005+..............+1/2004
=> a=b
Sper sa-i foloseasca vreunui studios cindva, undeva...:-)
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