已知數列{an}的前n項和為Sn.(1)若Sn=(-1)n+1·n,求a5+a6及an;(2)若Sn=3n+2...
問題詳情:已知數列{an}的前n項和為Sn.(1)若Sn=(-1)n+1·n,求a5+a6及an;(2)若Sn=3n+2n+1,求an.【回答】解:(1)因為a5+a6=S6-S4=(-6)-(-4)=-2,當n=1時,a1=S1=1,當n≥2時,an=Sn-Sn-1=(-1)n+1·n-(-1)n·(n-1)=(-1)n+1·[n+(n-1)]=(-1)n+1·(2n-1),又a1也適合此式,所以an=(-1...