设a=2008
则根号下=√a(a+1)(a+2)(a+3)+1=√(a^2+3a)(a^2+3a+2)+1=√(a^2+3a)^2+2(a^+3a)+1=a^2+3a+1
则P=a^2+3a+1-(a+1)^2=a=2008