问题: 高二数列2
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解答:
(1)
{an=n^2
{a1=1
a(n+1)+a(n+2)+……+a(2n)
=S(2n)-Sn
=1/6(2*2n+1)(2n+1)*2n-1/6*n(n+1)(2n+1)
=1/6(2n+1)[8n^2+2n-n^2-n]
=1/6(2n+1)(7n^2+n)
=1/16(2n+1)(7n+1)n
Sn=(2n+1)n
bn=Sn-S(n-1)=(2n+1)n-[(2(n-1)+1](n-1)
=2n^2+n-2n^2+4n-2-n+1
=4n-1
bn =4n-1
(2)
Tn/T(n-1)=2^[-(4n-1)/2^[-4(n-1)+1]
=2^[-4n+1+4n-4-1]
=2^(-4)
T1=2^[-4(-1)]=2^(-3)
T1(1-q^n)/(1-q)=2^(-3)[1-2^(-4n)]/[1-2^(-2)]>31/240
1/8(1-1/2^4n)/(1-1/16)>31/16*15
1-1/2^4n>(31/30)*15/16=31/32
32*2^4n-32>31*2^4n
2^4n>32
2^4n>2^5
n>5/4
n>=2
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