Question

一道关于高中数学的等比数列的题

数列{a的第n项}的前n项和计为Sn，已知a1=1，a的第(n+1)项=Sn*(n+2)/n

求证：(1)数列{Sn/n}是等比数列
(2)前n+1项之和,即S(n+1)=4*(a的第n项)

Answer 1

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)

Answer 2

a[n+1]=S[n]*(n+2)/n
a[1]=1
a[2]=1*(1+2)/1=3
S[n+1]=S[n]+a[n+1]=S[n]+S[n]*(n+2)/n=S[n]*(2n+2)/n
S[n+1]/(n+1)=S[n]*2/n
所以数列{Sn/n}是等比为2的等比数列

因为S[n]/n是以2为公比的等比数列，首项为S[1]/1=S[1]=A[1]=1
所以S[n]/n的通项公式是2^(n-1)
所以S[n ]= n*2^(n-1)
S[n-1]= (n-1)*2^(n-2)
所以A[n] = S[n ]- 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A[n] = (n+1) * 2^(n-2)
S[n+1]=(n+1)*2^n=4*A[n]

Answer 3

1. a(n+1)=S(n+1)-Sn=Sn*(n+2)/n
n*S(n+1)=(n+2)*Sn+n*Sn=2(n+1)*Sn
[S(n+1)/(n+1)]/[Sn/n]=2
数列{Sn/n}是等比数列 公比q=2 首项S1/1=S1=a1=1
2. Sn/n=(S1/1)*2^(n-1)
Sn=n*2^(n-1)
an=Sn-S(n-1)=n*2^(n-1)-(n-1)*2^(n-2)=(n+1)*2^(n-2)
an=(n+1)*2^(n-2)=(n+1)*2^n/4
(n+1)*2^n=4*an
S(n+1)=(n+1)*2^(n)=4*an