Question

这题怎么做 要详解

Answer 1

a1=1/2
2a(n+1)-an=n............(1)
2an-a(n-1)=n-1..........(2)
(1)-(2)得
2a(n+1)-3an+a(n-1)=1
2[a(n+1)-an]-[an+a(n-1)]=1
2[a(n+1)-an-1]=[an+a(n-1)-1]
bn=a(n+1)-an-1,则b(n-1)=an+a(n-1)-1
2bn=b(n-1)
bn/b(n-1)=1/2
{bn}是等比数列
2)a1=1/2,2a(n+1)-an=n.
2a2-a1=1
a2=3/4
bn=a(n+1)-an-1,
b1=a2-a1-1=3/4-1/2-1=-3/4
bn=b1*(1/2)^(n-1)=-3/4*(1/2)^(n-1)=-3*(1/2)^(n+1)
a(n+1)-an-1=-3*(1/2)^(n+1)
a(n+1)-an=1-3*(1/2)^(n+1)
累加得:

a(n+1)-a1=(a2-a1)+(a3-a2)+...+[a(n+1)-an]=n-3[(1/2)^2+(1/2)^3+...+(1/2)^(n+1)]
a(n+1)-a1=n-3[1/4(1-(1/2)^n]/(1-1/2)]
a(n+1)-a1=n-3[1/2(1-(1/2)^n]=n-3/2+3*(1/2)^(n+1)
a(n+1)=a1+(n-3/2)+3*(1/2)^(n+1)
a(n+1)=(n-1)+3*(1/2)^(n+1)
an=(n-2)+3*(1/2)^n

追问

能不能写在纸上

太麻烦了