Question

已知数列{an}的前n项和伟Sn，且a1=1，na（n+1）=（n+2）Sn，n属于N* 求证数列{Sn/n}为等比数列

求数列{an}的通项公式及前n项和Sn;
若数列{bn}满足：b1=1/2，bn+1/n+1=bn+Sn/n+1(n属于N*），求数列{bn}的通项公式

Answer 1

1） 证明：
na[n+1] = (n+2)S[n]
n(S[n+1]-S[n]) = (n+2)S[n]
nS[n+1] = 2(n+1)S[n]
S[n+1]/(n+1) = 2*S[n]/n, (首项=S[1]/1=a[1]/1=1)
所以：{S[n]/n}是以1为首项,公比为2的等比数列
2） 解：
S[n]/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)
= (2n-(n-1))* 2^(n-2) = (n+1)*2^(n-2)
3) 看不懂 加几个括号会死啊

追问

（3）若数列{bn}满足：b1=1/2，（bn+1）/（n+1）=bn+Sn/（n+1）(n属于N*），求数列{bn}的通项公式

Answer 2

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