Question

数列问题

设数列an=1/(n^2),bn=7;(n=1);=7/n-7/(n-1);(n≥2)；n属于正整数，构造函数Tn=(1-a2)(1-a3)…(1-an)，Pn=(1+b1)+(1+b2)+…+(1+bn)，则使Tn≤k*Pn对n≥2恒成立的k的最小值为______。

Answer 1

解：
Tn=(1-a2)(1-a3)...(1-an)
=(1- 1/2²)(1-1/3²)...(1-1/n²)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)...(1+1/n)(1-1/n)
=(3/2)(1/2)(4/3)(2/3)...[(n+1)/n][(n-1)/n]
={(1/2)(2/3)...[(n-1)/n]}{(3/2)(4/3)...[(n+1)/n]}
=[(1×2×...×(n-1)/(2×3×...×n)][3×4×...×(n+1)/(2×3×...×n)]
=(1/n)×[(n+1)/2]
=(n+1)/(2n)
Pn=(1+b1)+(1+b2)+...+(1+bn)
=n+b1+(b2+b3+...+bn)
=n+7+7[1/2-1/1+1/3-1/2+...+1/n-1/(n-1)]
=n+7+7(1/n -1)
=n+ 7/n
Tn≤kPn
(n+1)/(2n)≤k(n+7/n)
整理，得
k≥(n+1)/(2n²+14)
[(n+1)+1]/[2(n+1)²+14] -(n+1)/(2n²+14)
=(n+2)/(2n²+4n+16) -(n+1)/(2n²+14)
=[(n+2)(2n²+14)-(n+1)(2n²+4n+16)]/[(2n²+4n+16)(2n²+14)]
=(-2n²-6n+12)/[(2n²+4n+16)(2n²+14)]
=-2(n²+3n-6)/[(2n²+4n+16)(2n²+14)]
=-2[(n²+3n-4)-2]/[(2n²+4n+16)(2n²+14)]
=-2[(n+4)(n-1)-2]/[(2n²+4n+16)/(2n²+14)]
n≥2 n+4≥6 n-1≥1 (n+4)(n-1)-2≥4>0
-2(n²+3n-6)/[(2n²+4n+16)(2n²+14)]恒<0
即n≥2时，随n递增，(n+1)/(2n²+14)单调递减，当n=2时取得最大值，要不等式k≥(n+1)/(2n²+14)恒成立，则k应≥这个最大值。
k≥(2+1)/(2×2²+14)
k≥3/22
k的最小值为3/22。

Answer 2

表示题目没看懂