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2021年贵州省高考数学二轮解答题专项复习:数列
1.已知等比数列{a n }是递减数列,a 1a 4=
132,a 2+a 3=38
. (1)求数列{a n }的通项公式;
(2)若b n =﹣(n +1)log 2a n ,求数列{1b n }的前n 项和T n .
【解答】解:(1)等比数列{a n }是递减数列,设公比为q ,
a 1a 4=132,a 2+a 3=38,可得a 2a 3=132,
解得a 2=14,a 3=18,满足a 2>a 3,
解得a 1=q =12,则a n =(12)n ; (2)b n =﹣(n +1)log 2a n =﹣(n +1)?(﹣n )=n (n +1),
1
b n =1n(n+1)=1n ?1n+1
, 可得前n 项和T n =1?12+12?13+?+1n ?1n+1=1?1n+1=n n+1.
2.已知{a n }是公差为1的等差数列,数列{b n }满足b 1=1,b 2=12
,a n b n +1+b n +1=nb n .
(1)求数列{b n }的通项公式;
(2)设c n =b n b n +1,求数列{c n }的前n 项和S n .
【解答】解:(1)由题意,可知a 1b 2+b 2=b 1,
即12a 1+12=1,解得a 1=1. 又∵数列{a n }是公差为1的等差数列,
∴a n =1+n ﹣1=n .
∴a n b n +1+b n +1=(n +1)b n +1=nb n ,
∴数列{nb n }是常数数列,即nb n =1?b 1=1,
∴b n =1n ,n ∈N *.
(2)由(1)知,c n =b n b n +1=1n(n+1)=1n ?1n+1,
故S n =c 1+c 2+…+c n
=1?12+12?13+?+1n ?1n+1
=1?1n+1