Question

数列an的前n项和记为sn，已知a1=1 an+1=（n分之n+2）乘sn（n=1，2，3，...)求证

1.数列n分之sn是等比数列
2。sn+1=4an

Answer 1

a(n+1)=S(n+1)-Sn=(n+2)/n*Sn
nS(n+1)-nSn=(n+2)Sn
nS(n+1)=2(n+1)Sn
S(n+1)/(n+1)=2(Sn/n)
[S(n+1)/(n+1)]/(Sn/n)=2
{Sn/n}是以2为公比的等比数列

Answer 2

解：
(1) 由已知有：
S[n+1] = S[n]+a[n+1] = S[n] + (n+2)/n *S[n]
==> S[n+1]/(n+1) = 2S[n]/n
==> { S[n+1]/(n+1)}/{S[n]/n} =2
可知 {S[n]/n} 为公比为2的等比数列；
(2 S[1] =a[1] =1，因此
S[n]/n = (S[1]/1)* 2^(n-1) = 2^(n-1)
==> S[n] = n*2^(n-1)
a[n] = S[n] - S[n-1] = n*2^(n-1) -(n-1)*2^(n-2) = (n+1)*2^(n-2)
S[n+1] = (n+1)*2^n = 4*(n+1)*2^(n-2)
==> S[n+1] = 4a[n]
结论得证