题目

数列｛an｝中,a1=2,an+1=4an-3n+1,求数列Sn,证明不等式Sn+1

提问时间：2020-10-27

答案

a(n+1) = 4a(n) - 3n + 1,
a(n+1) - (n+1) = 4a(n) - 4n = 4[a(n) - n],
{a(n) - n}是首项为a(1)-1=1,公比为4的等比数列
a(n)-n=4^(n-1),
a(n) = n + 4^(n-1),n = 1,2,..
S(n) = a(1) + a(2) + ...+ a(n)
= 1 + 1 + 2 + 4 + ...+ n + 4^(n-1)
= 1 + 2 + ...+ n + 1 + 4 + ...+ 4^(n-1)
= n(n+1)/2 + [4^n - 1]/(4-1)
= n(n+1)/2 + (4^n - 1)/3
所以S(n+1) -4S(n)=(n+1)(n+2)/2+(4^(n+1)-1)/3-4[n(n+1)/2 + (4^n - 1)/3]
=-(n+1)(3n-2)/2+1≤0对任意n属于正整数成立
所以不等式Sn+1≤4Sn,对任意n属于正整数成立