由数列{an}满足a1=t,an+1-an+2=0(t∈N*,n∈N*),知数列{an}是首项为t,公差为-2的等差数列,由此能求出an和数列{an}的前n项和的最大值为f(t).
【解析】
∵数列{an}满足a1=t,an+1-an+2=0(t∈N*,n∈N*),
∴数列{an}是首项为t,公差为-2的等差数列,
∴an=t+(n-1)×(-2)=-2n+t+2.
∴数列{an}的前n项和
S=nt+=nt+n-n2=-[n2-(t+1)n]=-(n-)2+,
∵数列{an}的前n项和的最大值为f(t),
∴当t为偶数时,f(t)=-+=;
当t为奇数时,f(t)=.
故f(t)=.
故答案为:-2n+t+2,.