an一(一1)^a(n一1)=n,则数列前40数列an前n项和为sn

来源：蜘蛛抓取(WebSpider) 时间：2016-04-02 10:37 标签： 求数列an的前n项和tn
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等差数列数列练习题(一)教师版1
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你可能喜欢数列{an}满足an+an+1=（n∈N*），a2=2，Sn是数列{an}的前n项和，则S21为（　　）A. 5B. C. D.
由an+an+1=（n∈N*），a2=2，得1＝-32，2＝2，a3＝-32，a4＝2…，∴数列{an}的所有奇数项项为，所有偶数项为2，∴21＝a1+10S2＝72．故选：B．
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其他类似问题
由数列递推式依次求出数列的前几项，得到数列{an}的所有奇数项项为，所有偶数项为2，结合an+an+1=得答案．
本题考点：
数列递推式．
考点点评：
本题考查了数列递推式，关键是对数列规律的发现，是中档题．
扫描下载二维码一直数列｛an}的前n项和sn=a^2n-1(a不等于0,正负1；n属于N,试判断｛an)是否为等比数列,为什么?
Sn=a^(2n-1)S(n-1)=a^(2n-3)an=Sn-S(n-1)=a^(2n-1)-a^(2n-3)=a^(2n-3)*(a²-1）因为a不等于0,正负1所以an不等于0所以an/a（n-1）=a²,a1=S1=a所以an为首项为a,公比为a²的等比数列
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其他类似问题
n=1时，a1=s1=a^2-1；n>=2时，an=s(n)-s(n-1)=(1-1/a^2)*a^2n; n=1,an=a^2-1 =a1;公比q=a(n)/a(n-1)=a^2;所以an为首项为a^2-1，公比为a^2的等比数列
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