2an+1=an+an+2
a(n+1)-a(n)=a(n+2)-a(n+1)
所以 {an}是等差數(shù)列
所以 a1=1,d=a2-a1=1
所以 an=1+(n-1)=n
bn=1/an*a(n+1)=1/n(n+1)=1/n-1/(n+1)
前五項(xiàng)和=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1-1/6
=5/6