已知公差为d的等差数列的前n项之和为Sn,则Sn+1/n+1-Sn/n
设Sn为等差数列{an}的前n项和,若a1=1,公差d=2,Sn+2-Sn=36,则n=( )
已知等差数列{an}的前n项和为Sn,Sn=kn(n+1)-n(k∈R),公差d为2.
已知等差数列{an}的前n项和为Sn,且a1不等于0,求(n*an)/Sn的极限、(Sn+Sn+1)/(Sn+Sn-1)
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1 -(2n+1)Sn=4n²-1(n∈N*)
已知等差数列{an}的首项a1=1,公差d=1,前n项和为Sn,bn=1/Sn 求证:b1+b2+.+bn
已知等差数列{an}的首项a1=1,公差d=1,前n项和为Sn,bn=1Sn,
已知等差数列{an}的公差d>0,其前n项和为sn,满足S90,则n=多少时,sn最小
若等差数列{an}的首项为a1,公差为d,前n项的和为Sn,则数列(Sn/n)为等差数列,且通项
.已知数列的前n项之和为Sn=n2+3n,求证{an}为等差数列,若Sn=n2+3n+1呢?
已知数列{an}是首项为1,公差为2的等差数列,Sn(n属于N)是数列的前n项和,则lim下面为n到无穷 Sn/n^2减
设等差数列的公差d为2,前n项和为Sn,则lim(n→∞)(an^2-n^2)/Sn
已知非负等差数列{an}的公差d不为0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:1/Sn+1/S