已知数列{an}满足条件:a1=1,an+1=2an+1,n∈N*.(Ⅰ)求证:数列{an+1}为等比数列;(Ⅱ)若bn=(2n

2025-02-27 09:20:07
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回答1:

解答:(Ⅰ)证明:由题意得an+1+1=2(an+1),…(3分)
又a1+1=2≠0.                                                 …(4分)
所以数列{an+1}是以2为首项,2为公比的等比数列.                 …(5分)
(Ⅱ)解:由(1)知an+1=2?2n?1an2n?1,…(7分)
bn=(2n?1)2n
∴Tn=b1+b2+b3+…+bn
=1?2+3?22+5?23+…+(2n-1)2n
2Tn=1?22+3?23+5?24+…+(2n-1)2n+1…(8分)
错位相减得-Tn=2+2?22+2?23+2?24+…+2?2n-(2n-1)2n+1…(9分)
=2[
2(1?2n)
1?2
]?2?(2n?1)2n+1
=(3-2n)2n+1-6…(11分)
从而得Tn=(2n-3)2n+1+6…(12分)