数列(an}的前n项和为Sn,且满足an=-3Sn*Sn-1(n>=2),a1=1/3 1.证(1/Sn}是等差数列
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/06 08:47:22
数列(an}的前n项和为Sn,且满足an=-3Sn*Sn-1(n>=2),a1=1/3 1.证(1/Sn}是等差数列
数列(an}的前n项和为Sn,且满足an=-3Sn*Sn-1(n>=2),a1=1/3
1.证(1/Sn}是等差数列
2.令bn=2(1-n)an(n>=2)求证b2平方+b3平方+.+bn平方
数列(an}的前n项和为Sn,且满足an=-3Sn*Sn-1(n>=2),a1=1/3
1.证(1/Sn}是等差数列
2.令bn=2(1-n)an(n>=2)求证b2平方+b3平方+.+bn平方
1)S[n]-S[n-1] = -3S[n]*S[n-1]
两边除以 - S[n]*S[n-1]
1/S[n] -1/S[n-1] =3
2)S[n]=1/(3n)
a[n]= S[n] - S[n-1] =1/( 3(1-n)n )
b[n]=2/(3n)
因为1/(n^2) < 1/( (n-1)n ) =1/(n-1) - 1/n
b2平方+b3平方+.+bn平方
=4/9*(1/2^2 +1/3^2+...+1/n^2)
两边除以 - S[n]*S[n-1]
1/S[n] -1/S[n-1] =3
2)S[n]=1/(3n)
a[n]= S[n] - S[n-1] =1/( 3(1-n)n )
b[n]=2/(3n)
因为1/(n^2) < 1/( (n-1)n ) =1/(n-1) - 1/n
b2平方+b3平方+.+bn平方
=4/9*(1/2^2 +1/3^2+...+1/n^2)
已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列
已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列
已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an
设数列[an}的前n项和为Sn,已知a1=1且满足3Sn的平方=an(3Sn-1),1求{1/Sn}为等差数列 1.求{
已知数列an的前n项和为Sn,且满足Sn=Sn-1/2Sn-1+1,a1=1/2(1)求证:1/Sn是等差数列(2
已知数列An的前n项和为Sn.且满足an+2Sn*Sn-1=0=2>,a1=1/2,求证1/Sn是等差数列,求通项an的
已知数列an的前n项和为Sn,且an+2Sn*Sn-1=0,a1=1/2,求证1/SN是等差数列,求数列SN的的通项公式
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2)
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2),
设数列{an}的前n项和为Sn,满足2Sn=an+1-2^n+1+1,且a1,a2+5.a3成等差数列
数列an的前n项和为Sn.且满足a1=1.2Sn=(n+1)an
数列{an}满足a1=1,设该数列的前n项和为Sn,且Sn,Sn+1,2a1成等差数列.用数学归纳法证明:Sn=(2n-