若an=2^(n-1),那么sn=?
若数列an的通项公式an=48-2n,那么Sn取最大值时,n=?
设数列{an}的前n项和为Sn,若a1=1,Sn=2an+Sn+(n∈N+),则a6=
数列{an}首相a1≠0,前n相的和为sn,满足Sn+1=2sn+a1,那么2an/sn等于多少,n趋近正无穷
数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn
已知an=1/2n(n+1),求Sn
数列Sn=(3n+1)/2-(n/2)an
An=n×2^(n-1),求Sn
an=(2^n-1)n,求Sn
已知an=(2n+1)*3^n,求Sn
递推与数列问题设数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2*an-n(n-1),试写出Sn与Sn-1(n
已知等差数列an中,a1=1,前n项和Sn,若S(n+1)/Sn=(4n+2)/(n+1),求an
已知a1=1,Sn=n^2an 求:an及Sn