Question

va rog frumos ex 3 :((((​

Answer 1

Folosesti formula S=[(a1+an)*n]/2

2+4+6+....+2004 = 2(1+2+3+..+1002)

S1 = 1+2+3+..+2004 = [(1+2004)*2004]/2 = 2005*1002

S2 = 2(1+2+3+..+1002)= 2*[(1+1002)*1002]/2 = 1003*1002

(1+2+3+..+2004)/(2+4+6+..+2004) = (2x-1)/x

S1/S2 = (2x-1)/x

(2005*1002)/(1003*1002) = (2x-1)/x

2005/1003 =(2x-1)/x

2005x=1003*(2x-1)

2005x=2006x-1003

2006x-2005x = 1003

x = 1003

Answer 2

Răspuns:

se aplica formula lui Gauss

S=[n(n+1)]/2

1+2+3+..+2004=[2004(2004+1)/2]=2004/2·2005=1002·2005

2+4+6+...+2004=dam factor comun 2 2(1+2+3+...+1002)=2(1002·1003)/2=1002·1003

inlocuim in ecuatie

(1002·2005)/1002·1003=(2x-1)/x se simplifica 1002

2005/1003=(2x-1)/x se face produsul mezilor cu extremii

2005x=1003(2x-1)

2005x=2006x-1003

1003=2006x-2005x

x=1003