Question

已知等差數(shù)列{an}中,a2+a4=10,a5=9,數(shù)列{bn}中,b1=a1,bn+1=bn+an.

(1)求數(shù)列{an}的通項(xiàng)公式,寫出它的前n項(xiàng)和Sn.

(2)求數(shù)列{bn}的通項(xiàng)公式.

(3)若cn=,求數(shù)列{cn}的前n項(xiàng)和Tn.

 

Answer 1

(1) an=2n-1,Sn= n2 (2) bn=n2-2n+2 (3) Tn= =

【解析】(1)設(shè){an}的公差為d,由題意得a1=1,d=2,

所以an=2n-1,Sn=na1+d=n2.

(2)b1=a1=1,bn+1=bn+an=bn+2n-1,

所以b2=b1+1,b3=b2+3=b1+1+3,

bn=b1+1+3+…+(2n-3)=1+(n-1)2=n2-2n+2(n≥2).

又n=1時(shí)n2-2n+2=1=b1,

所以數(shù)列{bn}的通項(xiàng)公式為bn=n2-2n+2.

(3)cn===-,

Tn=c1+c2+…+cn=(-)+(-)+…+(-)=1-=.