已知数列｛an｝的前n项和为Sn=n^2+1，数列｛bn｝满足bn=2/(an)+1，前n项和为Tn，设Cn=T(2n+1)-Tn 已知数列{an}的前n项和为Sn=—n^2 +...
\u5df2\u77e5\u6570\u5217{an}\u7684\u524dn\u9879\u548c\u4e3aSn\uff0c\u8bbe\u6570\u5217{bn}\u6ee1\u8db3bn\uff1d2\uff08Sn\uff0b1−Sn\uff09Sn−n\uff08Sn\uff0b1\uff0b\u8bbe\u7b49\u5dee\u6570\u5217{a[n]}\u7684\u516c\u5dee\u4e3ad\uff0c\u5219a[n+1]=a[1]+nd\uff0cS[n]=na[1]+(n\uff08n-1\uff09/2)d\uff0c
\u7531b[n]\uff1d2\uff08S[n+1]−S[n]\uff09S[n]−n\uff08S[n+1]\uff0bS[n]\uff09\uff08n\u2208(N^*)\uff09\uff0c\u5f97b[n]\uff1d2a[n+1]S[n]−n\uff082S[n]\uff0ba[n+1]\uff09
\u53c8\u7531b[n]\uff1d0\uff0c\u5f97
2\uff08a[1]+nd\uff09[na[1]+(n\uff08n-1\uff09/2)d]−n[2na[1]+n\uff08n-1\uff09d\uff0ba[1]+nd]=0\u5bf9\u4e00\u5207n\u2208(N^*)\u90fd\u6210\u7acb
\u5373\u5bf9\u4e00\u5207n\u2208(N^*)\uff0c
\uff08dn+a[1]\uff09[d(n^2)+\uff082a[1]-d\uff09n]−n\uff08d(n^2)+2a[1]n\uff0ba[1]\uff09=0
(d^2)(n^3)+\uff082a[1]d-(d^2)\uff09(n^2)+a[1]d(n^2)+\uff082(a[1]^2)-a[1]d\uff09n-d(n^3)-2a[1](n^2)-a[1]n=0
(\uff08d^2)-d\uff09(n^3)+\uff083a[1]d-(d^2)-2a[1]\uff09(n^2)+\uff082(a[1]^2)-a[1]d-a[1]\uff09n=0\uff08\u4e24\u8fb9\u540c\u9664\u4ee5n\uff09
(\uff08d^2)-d\uff09(n^2)+\uff083a[1]d-(d^2)-2a[1]\uff09n+2(a[1]^2)-a[1]d-a[1]=0
\u4ece\u800c(d^2)-d=0\uff0c3a[1]d-(d^2)-2a[1]=0\uff0c2(a[1]^2)-a[1]d-a[1]=0
\u89e3\u5f97d=0\uff0ca[1]=0
\u6216d=1\uff0ca[1]=1
\u7ecf\u68c0\u9a8c\u7b26\u5408\u9898\u610f\u3002\u6240\u4ee5{a[n]}\u7684\u901a\u9879\u516c\u5f0f\u4e3aa[n]=0\u6216a[n]=n
Sn\u662f\u5173\u4e8en\u7684\u4e8c\u6b21\u51fd\u6570\uff0c\u90a3\u4e48an\u4e00\u5b9a\u662f\u7b49\u5dee\u6570\u5217\u3002
a1=-1+1=0;
S2=-2^2+2=-2 \u2192a2=-2
\u5219\u516c\u5deed=a2-a1=-2
bn/b=2^(an-a)=2^d=1/4, bn\u662f\u4ee51/4\u4e3a\u516c\u6bd4\u7684\u7b49\u6bd4\u6570\u5217\u3002
b1=2^a1=1;
lim\uff08b1+b2+...+bn)
=b1/(1-q)
=1/(1-1/4)=4/3
(1)
∵数列{an}满足前N项和sn=n平方+1
∴Sn=n^2+1
S(n-1)=(n-1)^2+1
An=Sn-S(n-1)
=n^2+1-[(n-1)^2+1]
=2n-1
A1=S1=2
Bn=2/An +1=2/(2n-1)+1=(2n+1)/(2n-1)
B1=2/A1+1=2
Bn是一个首项为2，通项为(2n+1)/(2n-1) 的数列
(2)
Cn=T(2n+1)-Tn
要判断Cn的单调性只要判断Cn-C(n-1)是大于0还是小于0即可
Cn-C(n-1)=T(2n+1)-Tn-[T(2n-1)-T(n-1)]
=[T(2n+1)-T(2n-1)]-[Tn-T(n-1)]
=B(2n+1)+B(2n)-Bn
=[2(2n+1)+1]/[2(2n+1)-1]+[2(2n)+1]/[2(2n)-1]-[(2n+1)/(2n-1)]
=1+2[1/(4n+1)+1/(4n-1)-1/(2n-1)]
∵1/(4n+1)+1/(4n-1)-1/(2n-1)
= (1-8n)/[(4n+1)*(4n-1)*(2n-1)]
又∵1-8n<0,4n+1>0,4n-1>0,2n-1>0
∴(1-8n)/[(4n+1)*(4n-1)*(2n-1)]<0
但1+2[1/(4n+1)+1/(4n-1)-1/(2n-1)]<0
Cn单调递减
(3)
Sn=n^2+1
Sn-1=(n-1)^2+1
∴an=2n-1
bn=2/(2n-1+1)=1/n
Cn=bn+1+...+b2n+1=1/(n+1)+1/(n+2)..+1/(2n+1)
Cn+1=1/(n+2)+...+1/(2n+1)+1/(2n+2)+1/(2n+3)
Cn-Cn+1=1/(n+1)-1/(2n+2)-1/(2n+3)=1/(2n+2)-1/(2n+2)+1/(2n+2)-1/(2n+3)=1/(2n+2)(2n+3)>0
∴Cn为递减函数
C2=38/60>16/21>C3=319/420
∴k=3
1. Sn-S(n-1)=An An=2n-1 Bn=2/(2n-1)+1
题真的错了。bn=2/（an+1）
的通项公式；（2）求证数列｛Cn｝是单调递减数列
题有错吧！怎么Cn证出来为增数列
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绛旓細An = Sn-S(n-1)= an^2+bn+c-a(n-1)^2-b(n-1)-c = a(n+n-1)(n-n+1)+b(n-n+1)= a(2n-1)+b 鎵浠 A(n-1) = a[2(n-1)-1]+b = a(2n-3)+b 鎵浠 An-A(n-1) = a(2n-1)+b-a(2n-3)-b = 2a 鏄父鏁 鎵浠鏁板垪{An}浠庣浜岄」璧凤紝鏄叕宸负2a鐨勭瓑宸...
绛旓細Sn=nan,S(n-1)=(n-1)a(n-1),鐩稿噺锛歛n=nan - (n-1)a(n-1),鈭(n-1)an=(n-1)a(n-1),an=a(n-1)锛屽嵆an=1
绛旓細a(n-1)=(n-2)a(n-2)/n 鈥︹︹2=a1/3 杩炰箻 a2a3...an=a1a2...a(n-1)[(n-1)(n-2)...1]/[(n+1)n...3]=2a1a2...a(n-1)/[n(n+1)]an=2a1/[n(n+1)]=2/[n(n+1)]n=1鏃讹紝a1=2/(1脳2)=1锛屽悓鏍锋弧瓒炽鏁板垪{an}鐨閫氶」鍏紡涓篴n=2/[n(n+1)]绗簩...
