请问:无穷等比数列{an}的前n项和是S=a1/(1-q)还是Sn=a1(1-q^n)/(1-q)?
请问:无穷等比数列{an}的前n项和是S=a1/(1-q)还是Sn=a1(1-q^n)/(1-q)?
等比数列{an}的首项a1=1,公比为q且满足q的绝对值小于1.,前n项和为Sn,各项之和为S,
等比数列{an}的首项a1=1,公比为q,前n项和是Sn,则数列{1an}的前n项和是( )
等比数列求和公式推导首项a1,公比q a(n+1)=an*q=a1*q^(n Sn=a1+a2+..+an q*Sn=a
若等比数列{an}的首项a1=1,公比为q,前n项和是Sn,则数列{1/an}的前n项和为
等比数列求和公式Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1)
已知等比数列{an}的首项a1=2,公比q=3,Sn是它的前n项和.求证:Sn+1/Sn
已知数列{an}是首项a1=4,公比q不等于1的等比数列,Sn是其前n项和,且4a1,a5,-2a3成等差数列
等比数列前n项和公式Sn=a1(1-q^n)/(1-q)怎样推到出Sn=(a1-anq)/(1-q)
已知数列{an}是首项为a1,公比为q(q>0)的等比数列,前n项和Sn,设Tn=Sn/S( n+1) (n=1,2,3
等比数列{an}的首项a1=-1,前n项和为Sn若S 10S 5=3132,则公比q等于( )
等比数列{an}中,a1+an=34,a2*a(n-1)=64,前n项和Sn=62,求项数n及公比q的值