Question

数列{an}中,a1=1,a2=2,a(n+2)-an=1+(-1)^n(n∈N*),则S100

如题

Answer 1

(1)a(n+2)=(an+a(n+1))/2
a(n+2)-a(n+1)=(an+a(n+1))/2-a(n+1)=-1/2(a(n+1)-an)
即b(n+1)=-1/2bn
所以{bn}为等比数列
(2)b1=a2-a1=1
所以bn=(-1/2)^(n-1)
a(n+1)=an+(-1/2)^(n-1)
an=a(n-1)+(-1/2)^(n-2)
……
a3=a2+(-1/2)
a2=a1+1
用累加法，得an=a1+1+(-1/2)+(-1/2)^2+...+(-1/2)^(n-2)
=1+[1-(-1/2)^(n-1)]/[1-(-1/2)]=5/3-2/3(-1/2)^(n-1)

追问

求前100项和S100