问题
填空题
设Sn=1+2+3+…+n,n∈N*,则函数f(n)=
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答案
由题意Sn=1+2+3+…+n=n(n+1) 2
∴f(n)=
=Sn (n+32)Sn+1
= n(n+1) 2 (n+32) × (n+2)(n+1) 2
=n (n+32) ×(n+2)
≤1 n+34+ 64 n
=1 34+16
等号当且仅当n=1 50
=8时成立64 n
故答案为1 50
设Sn=1+2+3+…+n,n∈N*,则函数f(n)=
|
由题意Sn=1+2+3+…+n=n(n+1) 2
∴f(n)=
=Sn (n+32)Sn+1
= n(n+1) 2 (n+32) × (n+2)(n+1) 2
=n (n+32) ×(n+2)
≤1 n+34+ 64 n
=1 34+16
等号当且仅当n=1 50
=8时成立64 n
故答案为1 50