数列an的前n项和为Sn,a1=1/4且Sn=Sn-1+an-1+1/2(n-1为下标)
数列bn满足b1=-119/4,3bn-bn-1=n 求an通项公式,证:数列bn-an是等比数列,bn前n项和Tn的最小值
数列an不一定是等差数列~~
人气:203 ℃ 时间:2020-01-29 15:22:30
解答
1
Sn=Sn-1 +an-1 +1/2
an-an-1=1/2
an=a1+(n-1)/2=1/4+(n-1)/2=n/2-1/4
2
3bn-bn-1=n
3bn-3n/2-3/4=bn-1-(n-1)/2 -1/4
设cn=bn-n/2-1/4
cn=bn-an
3cn=cn-1
所以数列bn-an是等比数列
b1=-119/4 ,c1=-119/4-1/2-1/4=-122/4=-61/2
cn=c1*(1/3)^(n-1) =(-61/2)*(1/3)^(n-1)
Scn=c1*(1-(1/3)^(n)/(1-1/3)=(-61/2)*(3/2)(1-1/3^n)
=(-183/4)(1-1/3^n)
San=(n/2)n/2=n^2/4
Tn=Sbn=Scn+San=n^2/4+(-183/4)(1-1/3^n)
n=1时,Tn最小=b1=-119/4题目中没有说数列an是等差数列~~Sn=Sn-1 +an-1 +1/2(Sn-Sn-1)-an-1=1/2an=Sn-Sn-1an-an-1=1/2an等差数列a1=1/4 an=1/4+(n-1)/2=n/2-1/4
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