Sn 1 Sn=2an-1 a1=3

a1=1,Sn=2an+1求Sna1=1,Sn=2an+1求Sna1=1,Sn=2an+1求Sna(n)=S(n)-S(n-1)=2*a(n)+1-(2*a(n-1)+1)=2*a(n)-2*a(n-

已知数列an首项a1不等于0,Sn+1=2Sn+a1,求极限（an/sn）已知数列an首项a1不等于0,Sn+1=2Sn+a1,求极限（an/sn）已知数列an首项a1不等于0,Sn+1=2Sn+a1

数列an中,a1=1,sn+1=sn-1/2an,则an=?数列an中,a1=1,sn+1=sn-1/2an,则an=?数列an中,a1=1,sn+1=sn-1/2an,则an=?s(n+1)=sn-

数列{an}满足Sn+Sn+1=5/3an+1,a1=4求an数列{an}满足Sn+Sn+1=5/3an+1,a1=4求an数列{an}满足Sn+Sn+1=5/3an+1,a1=4求anSn+S(n+

高二数学数列求和问题!在数列an中,a1=1,Sn为an前n项和,an=S(n-1)求Sn,an请帮我看看我分别先算Sn和an哪里出错了?1.先算Sn因为an=2Sn-1且an=Sn-Sn-1所以Sn

a1=1,Sn=n^2an,求ana1=1,Sn=n^2an,求ana1=1,Sn=n^2an,求anSn-S(n-1)=An,n^2An-(n-1)^2a(n-1)=An(n^2-1)An=(n-1

已知数列{an}满足:a1=3,an=Sn-1+2n,求数列an及sn已知数列{an}满足:a1=3,an=Sn-1+2n,求数列an及sn已知数列{an}满足:a1=3,an=Sn-1+2n,求数列

a1=1/2,Sn=n^2an-n(n-1),求Sn和ana1=1/2,Sn=n^2an-n(n-1),求Sn和ana1=1/2,Sn=n^2an-n(n-1),求Sn和an因an=Sn-S(n-1)

数列｛an｝,中,a1=1/3,设Sn为数列｛an｝的前n项和,Sn=n（2n-1）an求Sn数列｛an｝,中,a1=1/3,设Sn为数列｛an｝的前n项和,Sn=n（2n-1）an求Sn数列｛an｝

已知数列an的前n项和sn与通项an满足a1=2,sn+1sn=an+1,求sn已知数列an的前n项和sn与通项an满足a1=2,sn+1sn=an+1,求sn已知数列an的前n项和sn与通项an满足

Sn=n（2n-1）ana1=1/3求数列的Sn,anSn=n（2n-1）ana1=1/3求数列的Sn,anSn=n（2n-1）ana1=1/3求数列的Sn,an当n=1时,可知S1=a1=1/3.当

已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,Sn?已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,Sn?已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,S

已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,Sn?已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,Sn?已知a1=3,an=Sn-1+2^n(n大于等于2）,求an,S

a1=3,an=Sn-1+2n次方.n大于或等于2.求an和Sna1=3,an=Sn-1+2n次方.n大于或等于2.求an和Sna1=3,an=Sn-1+2n次方.n大于或等于2.求an和Sn为表达清

a1=3,an=Sn-1+2的n次方.n大于或等于2.求an和Sna1=3,an=Sn-1+2的n次方.n大于或等于2.求an和Sna1=3,an=Sn-1+2的n次方.n大于或等于2.求an和Sna

已知数列{an}a1=2前n项和为Sn且满足SnSn-1=3an求数列{an}的通项公式an已知数列{an}a1=2前n项和为Sn且满足Sn+Sn-1=3an求数列{an}的通项公式an已知数列{an

数列an满足a1=1/3,Sn=n(2n-1)an,求an数列an满足a1=1/3,Sn=n(2n-1)an,求an数列an满足a1=1/3,Sn=n(2n-1)an,求anSn=n(2n-1)an,

数列an中,a1=-2/3,sn+1/sn=an-2(n>1的整数）,求s1,s2,s3,猜想sn表达式,并证明数列an中,a1=-2/3,sn+1/sn=an-2(n>1的整数）,求s1,s2,s3

数列an前n项和为sn,如果a1=1,sn=3+2an+1,求数列通项公式和sn数列an前n项和为sn,如果a1=1,sn=3+2an+1,求数列通项公式和sn数列an前n项和为sn,如果a1=1,s

已知数列an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn已知数列an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn已知数列an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn由