记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/04/29 00:51:42
记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn
求证:数列(Sn/n)为等比数列
求数列(an)的通项公式
求证:数列(Sn/n)为等比数列
求数列(an)的通项公式
证明,
因为A(n+1) = (n+2)/n * Sn
所以Sn = n*A(n+1) / (n+2)
S(n-1) = (n-1)*An / (n+1)
所以An = Sn - S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1) * An
所以2n/(n+1) * An = n/(n+2) * A(n+1)
即A(n+1)/An = (2n+4)/(n+1)
所以(Sn/n) / (S(n-1)/(n-1)) = ( A(n+1)/(n+2) ) / ( An / (n+1))
= A(n+1)/An * (n+1)/(n+2)
= (2n+4)/(n+1) * (n+1)/(n+2) = 2
所以Sn/n是以2为公比的等比数列
(2)
因为Sn/n是以2为公比的等比数列,首项为S1/1=S1=A1=1
所以Sn/n的通项公式是2^(n-1)
所以Sn = n*2^(n-1)
S(n-1) = (n-1)*2^(n-2)
所以An = Sn - S(n-1) = n*2^(n-1) - (n-1)*2^(n-2)
= n*2^(n-1) - n*2^(n-2) + 2^(n-2)
= n*2^(n-2) + 2^(n-2)
= (n+1) * 2^(n-2)
当n=1时也满足,所以通项公式为An = (n+1) * 2^(n-2)
因为A(n+1) = (n+2)/n * Sn
所以Sn = n*A(n+1) / (n+2)
S(n-1) = (n-1)*An / (n+1)
所以An = Sn - S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1) * An
所以2n/(n+1) * An = n/(n+2) * A(n+1)
即A(n+1)/An = (2n+4)/(n+1)
所以(Sn/n) / (S(n-1)/(n-1)) = ( A(n+1)/(n+2) ) / ( An / (n+1))
= A(n+1)/An * (n+1)/(n+2)
= (2n+4)/(n+1) * (n+1)/(n+2) = 2
所以Sn/n是以2为公比的等比数列
(2)
因为Sn/n是以2为公比的等比数列,首项为S1/1=S1=A1=1
所以Sn/n的通项公式是2^(n-1)
所以Sn = n*2^(n-1)
S(n-1) = (n-1)*2^(n-2)
所以An = Sn - S(n-1) = n*2^(n-1) - (n-1)*2^(n-2)
= n*2^(n-1) - n*2^(n-2) + 2^(n-2)
= n*2^(n-2) + 2^(n-2)
= (n+1) * 2^(n-2)
当n=1时也满足,所以通项公式为An = (n+1) * 2^(n-2)
记数列(an)的前n项和为Sn已知a1=1,对任意n∈N*,均满足an+1=(n+2)/n)Sn
已知数列an满足a1=1,前n项的和为Sn 且对任意的n∈N*有(n+1)an-2Sn=3n-3
已知数列{an}满足a1=1,an+1=Sn+(n+1)(n∈N*),其中Sn为{an}的前n项和,
记数列{an}的前n项和为Sn,若3a(n+1)=3an+2 n属于N+,a1+a3+a5+...+a99=90
已知an=2^n-1/2^n,记数列{1/an}的前n项和为Sn,求证Sn<4/3.
已知数列an的前n项和为sn,且满足sn=n²an-n²(n-1),a1=1/2
已知数列{an}的前n项和为Sn,a1=-23,Sn+1Sn=an-2(n≥2,n∈N)
已知数列an的前n项和为Sn,又有数列bn,他们满足关系b1=a1,对于n∈N*,有an+Sn=n,b(n+1)=a(n
已知数列an的前n项和为Sn,且满足an+SnSn-1=0(n>=2,n∈N*),a1=1/2.
数列{an}的前n项和为Sn,a1=1,an+1=2Sn(n∈N*)
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}的首项a1=5,前n项和为Sn,且Sn+1=2Sn+n+5(n∈N*).