已知数列{an}满足:a1=1,且n为奇数时,an+1=2an,n为偶数时,an+1=an+1,n∈N*.(1)求a2,a3并证明数

2025-05-14 12:11:57
推荐回答(1个)
回答1:

(1)a2=2a1=2,a3=a2+1=3,
∵a2n+1=a2n+1=2a2n-1+1,
∴a2n+1+1=2(a2n-1+1),
∴数列{a2n-1+1}为公比是2的等比数列;
(2)S2n+1=(a1+a2)+(a3+a4)+…+(a2n-1+a2n)+a2n+1+a2n+1
=3a1+3a3+…+3a2n-1+a2n+1
由(1)知,
a2n?1+1=2n
a2n?12n?1
S2n+1=3[(2?1)+(22?1)+…+(2n?1)]+a2n+1=3(2

1?2n
1?2
?n)+2n+1?1=2n+3-3n-7