问题
填空题
考虑以下数列an,n∈N*:①an=n2+n+1;②an=2n+1;③an=ln
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答案
①an=n2+n+1 中
=n2+3n+4an+2+an 2
an+1=n2+3n+3
>an+1an+2+an 2
②an=2n+1中
=2n+3an+2+an 2
an+1=2n+3
=an+1an+2+an 2
③an=ln
,n n+1
=an+2+an 2
=ln(
•n+2 n+3
)n n+1 2
,ln n2+2n n2+4n+3 2
an+1=ln(
),2an+1=2ln(n+1 n+2
)=ln(n+1 n+2
),n2+2n+1 n2+4n+4
计算得
<an+1an+2+an 2
当数列为等差数列时取等号,取得最小值
所以:a1=1,a20=a1+(n-1)d=58
∴d=3
∴a10=a1+9d=28
∴a10的最小值为:28
故答案为:②③;28