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已知数列{an}中,a₁=14,a₂=2,且满足a‹n+2›-2a‹n+1›+a‹n›=0;
设b‹n›=1/[n(18-a‹n›)],T‹n›=b₁+b₂+…+b‹n›,是否存在最大整数m,对任意n∈N,
均有T‹n›>m/32.
由a‹n+2›-2a‹n+1›+a‹n›=0,得a‹n+2›-a‹n+›=a‹n+1›-a‹n›,故{a‹n›}是等差数列.设公差为d
则d=a₂-a₁=2-14=-12;故a‹n›=14-12(n-1)=-12n+26.
于是b‹n›=1/[n(18+12n-26)]=1/[n(12n-8)]=1/[4n(3n-2)]=(1/4)[1/n(3n-2)]
故T‹n›=(1/4)[1/(1×1)+1/(2×4)+1/(3×7)+1/(4×10)+.+1/n(3n-2)]
>(1/4)+(1/4){1/(2×4)+1/(4×6)+1/(6×8)+.+1/[2n×(2n+2)]
=(1/4)+(1/8){(1/2-1/4)+(1/4-1/6)+(1/6-1/8)+.+[1/2n-1/(2n+2)]}
=1/4+(1/8)[1/2-1/(2n-2)]=1/4+1/8-1/[16(n-1)]=3/8-1/[16(n-1)]
故可取3/8=m/32,即m=96/8=12.
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