数列{An}是等差数列,公差d>0,Sn是{An}的前项和,已知A1A4=4,S4=10 (1)求数列{An}的通项公式(2)令Bn=1/An(An+1),求数列{Bn}的前n项和Tn(3)求f(n)=Sn/(n+32)S(n+1)的最大值及取得最大值时n
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![数列{An}是等差数列,公差d>0,Sn是{An}的前项和,已知A1A4=4,S4=10 (1)求数列{An}的通项公式(2)令Bn=1/An(An+1),求数列{Bn}的前n项和Tn(3)求f(n)=Sn/(n+32)S(n+1)的最大值及取得最大值时n](/uploads/image/z/1761708-12-8.jpg?t=%E6%95%B0%E5%88%97%7BAn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%85%AC%E5%B7%AEd%3E0%2CSn%E6%98%AF%7BAn%7D%E7%9A%84%E5%89%8D%E9%A1%B9%E5%92%8C%2C%E5%B7%B2%E7%9F%A5A1A4%3D4%2CS4%3D10+%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E4%BB%A4Bn%3D1%2FAn%28An%2B1%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7BBn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%EF%BC%883%EF%BC%89%E6%B1%82f%EF%BC%88n%EF%BC%89%EF%BC%9DSn%EF%BC%8F%EF%BC%88n%EF%BC%8B32%EF%BC%89S%EF%BC%88n%EF%BC%8B1%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%8F%8A%E5%8F%96%E5%BE%97%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6n)
数列{An}是等差数列,公差d>0,Sn是{An}的前项和,已知A1A4=4,S4=10 (1)求数列{An}的通项公式(2)令Bn=1/An(An+1),求数列{Bn}的前n项和Tn(3)求f(n)=Sn/(n+32)S(n+1)的最大值及取得最大值时n
数列{An}是等差数列,公差d>0,Sn是{An}的前项和,已知A1A4=4,S4=10 (1)求数列{An}的通项公式
(2)令Bn=1/An(An+1),求数列{Bn}的前n项和Tn
(3)求f(n)=Sn/(n+32)S(n+1)的最大值及取得最大值时n的值
数列{An}是等差数列,公差d>0,Sn是{An}的前项和,已知A1A4=4,S4=10 (1)求数列{An}的通项公式(2)令Bn=1/An(An+1),求数列{Bn}的前n项和Tn(3)求f(n)=Sn/(n+32)S(n+1)的最大值及取得最大值时n
由于{an}为等差数列,因此有:
a1.a4=a1(a1+3d)=(a1)^2+3a1.d=4
S4=4a1+6d=10
由于 d>0,因此代入解得:
d=1 ,a1=1
因此{an}通项公式为:
an=a1+(n-1)d
=n (n属于N+)
(2)
Bn=1/[an.a(n+1)]
1/[n(n+1)] (n属于N+)
Tn=B1+B2+B3+……+BN=1/(1x2)+1/(2x3)+1/(3x4)+……+1/[n(n+1)]
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1/n-1/(n+1)]
=1-1/(n+1)
=n/(n+1) (n属于N+)
(3)
由(1)可得:
Sn=na1+n(n-1)d/2=n+n(n-1)/2
=(n^2+n)/2 (n属于N+)
S(n+1)=(n^2+3n+2)/2 (n属于N+)
f(n)=Sn/[(n+32)S(n+1)] =[(n^2+n)/2] /[(n+32)(n^2+3n+2)/2]
=(n^2+n) /[(n+32)(n^2+3n+2)]
=(n^2+n) /[(n+32)(n+1)(n+2)]
=n /[(n+32)(n+2)]
=1/[n+64/n+34] (n属于N+)
由于n>0,故要使f(n)取最大值,则分母[n+64/n+34]取最小值,故由均值不等式可得:
[n+64/n+34] ≥2√[n(64/n)] +34 = 50
所以:
f(n)≤ 1/50
当取等号时,有:
n=64/n
解得:
n=8
故当n=8时,f(n)取得最大值 1/50
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