Question

求学霸帮忙解题！

Answer 1

2nS(n+1)-2(n+1)Sn=n(n+1)

等式两边同除以:2n(n+1)得
S(n+1)/(n+1)-Sn/n=1/2
{Sn/n}是等差数列
Sn/n=S1+1/2(n-1)=1+1/2(n-1)=(n+1)/2
Sn=n(n+1)/2
S(n-1)=n(n-1)/2
an=Sn-S(n-1)=n(n+1)/2-n(n-1)/2=n
an=n,n=1时a1=1成立
an=n
b(n+2)-b(n+1)=b(n+1)-bn
{bn}是等差数列
b3=b1+2d=5,S9=9b1+36d=63
解得:b1=3,d=1
bn=3+(n-1)=n+2
bn=n+2
即,an=n,bn=n+2
2)Cn=bn/an+an/bn=(n+2)/n+n/(n+2)=