已知数列{an}的前n项的和为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
\u5df2\u77e5\u6570\u5217{an}\u7684\u524dn\u9879\u548c\u4f1fSn\uff0c\u4e14a1=1\uff0cna\uff08n+1\uff09=\uff08n+2\uff09Sn\uff0cn\u5c5e\u4e8eN* \u6c42\u8bc1\u6570\u5217{Sn/n}\u4e3a\u7b49\u6bd4\u6570\u52171\uff09 \u8bc1\u660e\uff1a
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, (\u9996\u9879=S[1]/1=a[1]/1=1)
\u6240\u4ee5\uff1a{S[n]/n}\u662f\u4ee51\u4e3a\u9996\u9879,\u516c\u6bd4\u4e3a2\u7684\u7b49\u6bd4\u6570\u5217
2\uff09 \u89e3\uff1a
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) \u770b\u4e0d\u61c2 \u52a0\u51e0\u4e2a\u62ec\u53f7\u4f1a\u6b7b\u554a
\u8bc1\u660e\uff1a\u7531\u5df2\u77e5\u5f97\uff1an(S(n+1)-Sn)=(n+2)Sn\uff0c\u5373S(n+1)/(n+1)=2*Sn/n,\u6240\u4ee5\u6570\u5217Sn/n\u662f\u7b49\u6bd4\u6570\u5217\uff0c\u6240\u4ee5Sn=n*2^(n-1) \uff0cTn=1*2^0+2*2^1+3*2^2+...+n*2^(n-1) \uff0c 2Tn=\u3002\u3002\u3002\u76f8\u51cf\u53ef\u5f97\uff1aTn=(n-1)*2^n+1
1. na(n+1)=n[S(n+1)-Sn]=(n+2)SnnS(n+1)=2(n+1)Sn
S(n+1)/(n+1)=2*Sn/n
所以{Sn/n}是公比为2的等比数列
2. S1/1=a1=1
所以Sn/n=2^(n-1)
Sn=n*2^(n-1)
所以na(n+1)=(n+2)*n*2^(n-1)
a(n+1)=(n+2)*2^(n-1)
an=(n+1)*2^(n-2)
3. b(n+1)/(n+1)=[bn+n*2^(n-1)]/n
所以
b(n+1)/(n+1)-bn/n=2^(n-1)
bn/n-b(n-1)/(n-1)=2^(n-2)
....
b2/2-b1=2^0=1
叠加 b(n+1)/(n+1)-b1=1+2+2^2+...+2^(n-1)=2^n-1
b(n+1)=(n+1)(2^n-1)
故bn=n*2^(n-1)-n
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1. na(n+1)=n[S(n+1)-Sn]=(n+2)Sn
nS(n+1)=2(n+1)Sn
S(n+1)/(n+1)=2*Sn/n
所以{Sn/n}是公比为2的等比数列
2. S1/1=a1=1
所以Sn/n=2^(n-1)
Sn=n*2^(n-1)
所以na(n+1)=(n+2)*n*2^(n-1)
a(n+1)=(n+2)*2^(n-1)
an=(n+1)*2^(n-2)
3. b(n+1)/(n+1)=[bn+n*2^(n-1)]/n
所以
b(n+1)/(n+1)-bn/n=2^(n-1)
bn/n-b(n-1)/(n-1)=2^(n-2)
....
b2/2-b1=2^0=1
叠加 b(n+1)/(n+1)-b1=1+2+2^2+...+2^(n-1)=2^n-1
b(n+1)=(n+1)(2^n-1)
故bn=n*2^(n-1)-n
na(n+1)=n[s(n+1)-s(n)]=(n+2)s(n),
ns(n+1)=2(n+1)s(n),
s(n+1)/(n+1)=2s(n)/n,
s(n+1)/[(n+1)2^(n+1)] = s(n)/[n*2^n] = ... = s(1)/[1*2] = a(1)/2 = 1/2.
s(n)=(1/2)n2^n=n*2^(n-1).
s(n)/n = 2^(n-1)
{s(n)/n}是首项为1, 公比为2的等比数列.
na(n+1)=(n+2)s(n)=(n+2)*n*2^(n-1),
a(n+1)=(n+2)*2^(n-1),
a(n)=(n+1)2^(n-2).
b(n+1)/(n+1)=[b(n)+s(n)]/n=b(n)/n + 2^(n-1),
c(n)=b(n)/n,
c(n+1)=c(n)+2^(n-1),
c(n+1)/2^n = (1/2)c(n)/2^(n-1) + (1/2),
c(n+1)/2^n - 1 = (1/2)c(n)/2^(n-1) - 1/2 = (1/2)[c(n)/2^(n-1) - 1]
{c(n)/2^(n-1) - 1}是首项为c(1)-1=b(1)-1=-1/2,公比为（1/2）的等比数列。
c(n)/2^(n-1) - 1 = (-1/2)(1/2)^(n-1) = -1/2^n,
c(n) = 2^(n-1) - 1/2 = b(n)/n,
b(n) = n[2^(n-1) - 1/2]
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