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
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![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](/uploads/image/z/13336760-56-0.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%3D2n%26%23178%3B%2Bbn%2C%E4%B8%94a4%3D13%2C%E8%8B%A5%E6%95%B0%E5%88%97%EF%B9%9Bbn%EF%B9%9C%E6%BB%A1%E8%B6%B3b1%3D5%2Cbn%2B1%3Dabn%2C%E6%B1%82%EF%B9%9Bbn%EF%B9%9C%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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
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
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
假设an公差为d,则由于a4=13,所以an=(n-4)d+13
Sn=[(-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.因为前面的条件就已经能求出bn的通项了.后面给的 反而和前面的条件矛盾了.
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