高中数学数列题
\u9ad8\u4e2d\u6570\u5b66\u6570\u5217\u98981.
a(n+1)=(n+1)an/(2n)
a(n+1)/(n+1)=(1/2)(an/n)
[a(n+1)/(n+1)]/(an/n)=1/2\uff0c\u4e3a\u5b9a\u503c
a1/1=(1/2)/1=1/2\uff0c\u6570\u5217{an/n}\u662f\u4ee51/2\u4e3a\u9996\u9879\uff0c1/2\u4e3a\u516c\u6bd4\u7684\u7b49\u6bd4\u6570\u5217
an/n=1/2ⁿ
an=n/2ⁿ
\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3aan=n/2ⁿ
2.
Sn=a1+a2+...+an=1/2+2/2²+3/2³+...+n/2ⁿ
Sn /2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn -Sn /2=Sn /2=1/2+1/2²+...+1/2ⁿ -n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)
=1- (n+2)/2^(n+1)
Sn=2- (n+2)/2ⁿ
n=1\u65f6\uff0cb1=1\u00d7(2-S1)=1\u00d7(2-a1)=1\u00d7(2-1/2)=3/2
n=2\u65f6\uff0cb2=1\u00d7(2-S2)=1\u00d7(2-2+4/4)=1
n\u22652\u65f6\uff0c
bn=n(2-Sn)=n[2-2+(n+2)/2ⁿ]=n(n+2)/2ⁿ
b(n+1)/bn=[(n+1)(n+3)/2^(n+1)]/[n(n+2)/2ⁿ]
=(n+1)(n+3)/[2n(n+2)]
=(n²+4n+3)/(2n²+4n)
n\u4e3a\u6b63\u6574\u6570\uff0cn²+4n+3>0 2n²+4n>0
2n²+4n-(n²+4n+3)=n²-3 n\u22652\uff0cn²-3>0 2n²+4n>n²+4n+3
0<(n²+4n+3)/(2n²+4n)<1\uff0c\u5373\u6570\u5217\u4ece\u7b2c2\u9879\u5f00\u59cb\u5355\u8c03\u9012\u51cf\uff0c\u53c8b2=1<b1\uff0c\u6570\u5217{bn}\u5355\u8c03\u9012\u51cf
\u8981bn\u2265\u03bb\u6070\u67094\u4e2a\u5143\u7d20\uff0c\u53ea\u8981b4\u2265\u03bb b5<\u03bb
b4=4\u00d7(4+2)/2⁴=3/2\u2265\u03bb \u03bb\u22643/2
b5=5\u00d7(5+2)/2^5=35/3235/32
\u7efc\u4e0a\uff0c\u5f97 35/32<\u03bb\u22643/2
S1=a1(1-q)/(1-q),S2=a1(1-q^2)/(1-q),...,Sn=a1(1-q^n)/(1-q).
S1+S2+...+Sn=[a1/(1-q)]*[1-q+1-q^2+...+1-q^n]
=[a1/(1-q)]*[n-q(1-q^n)/(1-q)]
=na1/(1-q)-a1q(1-q^n)/(1-q)^2,
nS=na1/(1-q),
(S1+S2+...+Sn)-nS=-a1q(1-q^n)/(1-q)^2,
lim[(S1+S2+...+Sn)-nS]=-a1q/(1-q)^2=q/(1-q)^2
\u795d\u60a8\u5b66\u4e60\u6109\u5feb
an-1=1-1/a(n-1) =[a(n-1)-1]/a(n-1)
两边取倒数得到
1/[an-1]=a(n-1)/[a(n-1)-1]=1+1/[a(n-1)-1]
也就是bn=1+b(n-1)
所以bn是等差数列
b1=1/(a1-1)=-5/2
所以bn=-5/2+1(n-1)=n-7/2
即1/(an-1) =n-3.5
所以an-1=1/(n-3.5)
所以an=1+1/(n-3.5)=1+10/(10n-35)
10/(10n-35)在(1,3)递减，（4，正无穷）递减
又a1=1+10/(10-35)=3/5
a3=1+10/(30-35)=-1
a4=1+10/(40-35)=3
n>4时an=1+10/(10n-35)>0
所以
最大项a4=3
最小项a3=-1
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