高中数学.数列.求第二问.

1。a2=(1/2)(a1+1)=(1/2)(1/2 +1)=3/4b1=a2-a1-1=3/4 -1/2 -1=-3/4bn=a(n+1)-an-1=(1/2)(an +n)-an-1=(-1/2)an +(n-2)/2b(n+1)=a(n+2)-a(n+1)-1=(1/2)[a(n+1)+(n+1)]-a(n+1)-1=(-1/2)a(n+1) +(n-1)/2=(-1/2)[(1/2)(an+n)]+(n-1)/2=(-1/4)an +(n-2)/4=(1/2)[(-1/2)an +(n-2)/2]b(n+1)/bn=(1/2)[(-1/2)an+(n-2)/2]/[(-1/2)an+(n-2)/2]=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。a(n+1)-an-1=bn=(-3/4)×(1/2)^(n-1)=-3/2^(n+1)a(n+1)-an=1- 3/2^(n+1)an-a(n-1)=1- 3/2ⁿa(n-1)-a(n-2)=1-3/2^(n-1)…………a2-a1=1- 3/2^2累加an-a1=(n-1)-3×(1/2^2+1/2^3+。。。+1/2ⁿ)an=a1+n-1 -3×(1/2^2+1/2^3+。。。+1/2ⁿ)=1/2 +n -1 -3×(1/4)×[1-(1/2)^(n-1)]/(1-1/2)=n +3/2ⁿ -2n=1时,a1=1+3/2-2=1/2,同样满足通项公式数列{an}的通项公式为an=n +3/2ⁿ -2