2a(n+1)=an +n2a(n+1)-2(n+1)+4=an-n+22[a(n+1)-(n+1)+2]=an-n+2[a(n+1)-(n+1)+2]/(an-n+2)=1/2,为定值a1-1+2=1/2-1+2=3/2,数列{an-n+2}是以3/2为首项,1/2为公比的等比数列an-n+2=(3/2)(1/2)^(n-1)=3/2ⁿan=n +...
2a(n+1)=an +n2a(n+1)-2(n+1)+4=an-n+22[a(n+1)-(n+1)+2]=an-n+2[a(n+1)-(n+1)+2]/(an-n+2)=1/2,为定值a1-1+2=1/2-1+2=3/2,数列{an-n+2}是以3/2为首项,1/2为公比的等比数列an-n+2=(3/2)(1/2)^(n-1)=3/2ⁿan=n +...