Question

Va rog urgeent de tot !!!
determinati n apartinand numerelor naturale nenule astfel incat : 1/1x3 +1/3x5+1/5x7+....+1/n(n+2)=1002/2005

Answer 1

1/1x3 +1/3x5+1/5x7+....+1/n(n+2)=1002/2005
1/(1x3)=1/2 ×(1/1-1/3)
1/2×[1/1-1/3+1/3-1/5+1/5-1/7+....+1/n-1/(n+2)]=1002/2005
1/2×[1/1-1/(n+2)]=1002/2005
(n+2-1)/(n+2)=1002*2/2005
(n+1)/(n+2)=2004/2005
2004(n+2)=2005(n+1)
2004n+4008=2005n+2005
n=4008-2005=2003