数列an中an=2n^2+n-1/2n^2+3n+1求证an属于(0,1) 5
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证:
an=(2n²+n-1)/(2n²+3n+1)
=(2n²+3n+1-2n-2)/(2n²+3n+1)
=1- 2(n+1)/(2n²+3n+1)
=1- 2(n+1)/[(n+1)(2n+1)]
=1- 2/(2n+1)
n∈N*,2/(2n+1)>0,1- 2/(2n+1)<1
随n增大,2n+1单调递增,2/(2n+1)单调递减,1- 2/(2n+1)单调递增
n=1时, 1-2/(2n+1)有最小值: 1- 2/(2×1+1)=⅓
n→+∞时,2/(2n+1)→0,1- 2/(2n+1)→1
综上,得:⅓≤1- 2/(2n+1)<1
0<⅓≤an<1
an∈(0,1)
从解题过程知,其实an的取值区间还可以进一步确定,是[⅓,1)
an=(2n²+n-1)/(2n²+3n+1)
=(2n²+3n+1-2n-2)/(2n²+3n+1)
=1- 2(n+1)/(2n²+3n+1)
=1- 2(n+1)/[(n+1)(2n+1)]
=1- 2/(2n+1)
n∈N*,2/(2n+1)>0,1- 2/(2n+1)<1
随n增大,2n+1单调递增,2/(2n+1)单调递减,1- 2/(2n+1)单调递增
n=1时, 1-2/(2n+1)有最小值: 1- 2/(2×1+1)=⅓
n→+∞时,2/(2n+1)→0,1- 2/(2n+1)→1
综上,得:⅓≤1- 2/(2n+1)<1
0<⅓≤an<1
an∈(0,1)
从解题过程知,其实an的取值区间还可以进一步确定,是[⅓,1)
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