Răspunsuri

2014-11-08T19:46:50+02:00
In general suma de numere impare este patrat perfect. Iata cazul general:
1=2*1-1
3=2*2-1
5=2*3-1
....
n=2*k-1
1+3+5+....+(2*k-1)=2*(1+2+3+...+k)-1*k=2*k*(k+1)/2-k=k*(k+1)-k=k*k+k-k=k*k, adica patrat perfect.

In cazul nostru, 2011=2*1006-1, deci k=1006

Aplici in formula de mai sus si gata rezolvarea.
1+3+5...+2011=
1+3+5....+2011= 1+2+3+4+5+6+7+8+9+....+2010-2011-2012= 1+3+5+7+....+2011=S2012-(2+4+6+....+2010+2012)=2012*2013:2-2(1+2+3+.....+1006)=1006*2013-2*1006*1007:2=1006*2013-1*1006*1007=1006(2013-1007)=1006*1006=1006 la puterea 2=Patrat perfect
2014-11-08T19:58:06+02:00
1= 1 +2·0
3 =1 +2·1
5 =1+2·2
7= 1+2·3
.............
2011 = 1+2·1005 ⇒ b= 1·1006 +2(1+2+3+......+1005) = 1006 +2·1005·1006/2 =
= 1006 (1 +1005) =1006² =p.p.
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