等差数列﹛an﹜的前n项和为Sn=2n²+bn,且a4=13,若数列﹛bn﹜满足b1=5,bn+1=abn,求﹛bn﹜的通项公式
问题描述:
等差数列﹛an﹜的前n项和为Sn=2n²+bn,且a4=13,若数列﹛bn﹜满足b1=5,bn+1=abn,求﹛bn﹜的通项公式
答
假设an公差为d,则由于a4=13,所以an=(n-4)d+13Sn=[(-3d+13)+(n-4)d+13]n/2=[(n-7)d+26]n/2=d/2×n²+(26-7d)/2 ×n所以d/2=2,bn=(26-7d)/2 ×n=-n.对于后面的若数列﹛bn﹜满足b1=5,bn+1=abn,我很不解....