已知数列{an}的前N项和为Sna1=1/4Sn=Sn-1+an-1+1/2数列{bn}满足3bn-b(n-1)=n求an通项求证{bn-an}为等求an通项求证{bn-an}为等比求bn前N项和Tn的最小值速求!

Question

已知数列{an}的前N项和为Sn a1=1/4 Sn=Sn-1+an-1+1/2 数列{bn}满足3bn-b(n-1)=n 求an通项 求证{bn-an}为等
求an通项 求证{bn-an}为等比 求bn前N项和Tn的最小值 速求!

Answer 1

1.
Sn=S(n-1)+a(n-1)+1/2
Sn-S(n-1)=a(n-1)+1/2
an=a(n-1)+1/2
an-a(n-1)=1/2
an=a1+(1/2)(n-1)
=1/4+(1/2)(n-1)
=n/2-1/4
2.设cn=bn-an,c1=b1-a1=b1-1/4
bn=cn+an=cn+n/2-1/4
3bn-b(n-1)=n
3(cn+n/2-1/4)-[c(n-1)+(n-1)/2-1/4]=n
3cn-c(n-1)=0
cn/c(n-1)=1/3
所以bn-an是首项b1-1/4公比为1/3的等比数列.
3.
bn-an=(b1-1/4)(1/3)^(n-1)
bn=n/2-1/4+(b1-1/4)(1/3)^(n-1)
Tn=[1/2-1/4+(b1-1/4)(1/3)^0]+[2/2-1/4+(b1-1/4)(1/3)^1]+[3/2-1/4+(b1-1/4)(1/3)^2]+……+[(n-1)/2-1/4+(b1-1/4)(1/3)^(n-2)]+[n/2-1/4+(b1-1/4)(1/3)^(n-1)]
=[1+2+3+……+(n-1)]/2-n/4+(b1-1/4)[(1/3)^0+(1/3)^1+(1/3)^2+……+(1/3)^(n-2)+(1/3)^(n-1)]
=n(n-1)/2-n/4+(b1-1/4)[1-(1/3)^n]/(1-1/3)
=n(2n-3)/4+(3/2)(b1-1/4)[1-(1/3)^n]
b1=?
再用导数求最大、最小值.