在等差数列中,a2=4,a5=10求证(1)1/Sn=1/n-1/(n+1) (2)1/S1+1/S2+.+1/Sn
sn=n^2 求证1/s1+1/s2+1/s3……1/sn
等差数列a1+a2+a3+a4=24 a5+a6+a7+a8=56求1/S1+1/S2+.1/Sn
已知Sn=1/2n(n+1),Tn=S1+S2+S3+.+Sn,求Tn.
An=2n-1,求证1/s1+1/s2+1/s3+…+1/sn
已知等差数列a[n]通项公式为a[n]=n,Sn是an^2和an的等差中项.求证1/S1+1/S2+...+1/Sn
等差数列前n项和Sn=(-1)^nAn-1/2^n 求 S1+S2+S3+……+S100
设数列an的前n项和为Sn,已知S1=1,Sn+1/Sn=n+c/n,且a1,a2,a3成等差数列
设数列{an}前n项和为Sn,已知(1/S1)+(1/S2)+.+(1/Sn)=n/(n+1),求S1,S2及Sn
已知数列an的前项和为Sn,a1=1,nSn+1-(n+1)Sn=n^2+cn,S1,S2/2,S3/3成等差数列.(1
设sn为数列an的前n项和,Sn=(-1)^n-1/2^n,n属于N*,则(1)a3=? (2)S1+S2+...+S1
设an=(1/n)*sin(nπ/25),Sn=a1+a2+…+an,在S1,S2,…,S100中,正数的个数是?(要详
已知数列{an}中,a1=1,且Sn,Sn+1,2S1成等差数列,用数学归纳法证明Sn=(2^n-1)/2^(n-1)