已知数列｛an｝中,a1=1/2,且2a(n+1)-an=n,其中n=1,2,3… 若bn=a(n+1)-an-1

1、证：2a(n+1)-an=n2a(n+1)=an+n2a(n+1)-2(n+1)+4=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 +3/2ⁿ -2a2=2+3/4 -2=3/4b1=a2-a1-1=3/4 -1/2 -1=-3/4bn=a(n+1)-an -1=(n+1)+3/2^(n+1) -2 -n -3/2ⁿ+2 -1=3/2^(n+1)b(n+1)/bn=[3/2^(n+2)]/[3/2^(n+1)]=1/2,为定值。数列{bn}是以-3/4为首项,1/2为公比的等比数列。2、由第一问得an=n+3/2ⁿ-2