在数列{an}中，a1=2，an+1=4an-3n+1，n∈N*．（Ⅰ）证明数_爱下问搜题

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问题解答题

在数列{a_n}中，a₁=2，a_n+1=4a_n-3n+1，n∈N^*．

（Ⅰ）证明数列{a_n-n}是等比数列；

（Ⅱ）求数列{a_n}的前n项和S_n；

（Ⅲ）证明不等式S_n+1≤4S_n，对任意n∈N^*皆成立．

答案

（Ⅰ）证明：由题设a_n+1=4a_n-3n+1，得a_n+1-（n+1）=4（a_n-n），n∈N^*．

又a₁-1=1，所以数列{a_n-n}是首项为1，且公比为4的等比数列．

（Ⅱ）由（Ⅰ）可知a_n-n=4^n-1，于是数列{a_n}的通项公式为a_n=4^n-1+n．

所以数列{a_n}的前n项和Sn=

4n-1

3

+

n(n+1)

2

．

（Ⅲ）证明：对任意的n∈N^*，Sn+1-4Sn=

4n+1-1

3

+

(n+1)(n+2)

2

-4(

4n-1

3

+

n(n+1)

2

)=-

1

2

(3n2+n-4)≤0．

所以不等式S_n+1≤4S_n，对任意n∈N^*皆成立．

完形填空

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