设sn=1-2+4-8+…+(-2)^(n-1) n ∈N*,S8等于
设Sn为等差数列{an}的前n项和,已知a9=-2,S8=2 (1)求首项和公差 (2)当n为何值,Sn最大?
若数列an满足:a1=1,Sn-1=2an+Sn(n∈N+),求an前8项和S8
Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗?
设数列{an}的前n项和为sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*)
已知Sn=2+5n+8n^2+…+(3n-1)n^n-1(n∈N*)求Sn
设Sn为等差数列{an}的前n项和,S8=4a3,a7=-2,a9=?
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*)
设Sn为数列{an}的前n项和,Sn=(-1)^n an - 1/(2^n),n∈N*,则 (1)a3=___ (2)S
设数列an的首项a1等于1,前n项和为sn,sn+1=2n
设数列an的前n项和为Sn,a1=1,an=(Sn/n)+2(n-1)(n∈N*) 求证:数列an为等差数列,
已知数列{an}的通项公式an=log2[(n+1)/(n+2)](n∈N),设其前n项的和为Sn,则使Sn
设数列{an}的前n项和Sn=(-1)^n(2n^2+4n+1)-1