a(n+1)-an=-2常数所以{an}为公差为-2的等差数列an=t-2(n-1)前n项和Sn=tn+(n-1)n/2*(-2)=-[n-(t+1)/2]^2+(t+1)^2/4显然,当n越接近(t+1)/2时,Sn的值越大因为t,n均为正整数当t为奇数时,(t+1)/2为整数,n=(t+1)/2时取得最...
已知数列{an}满足a1=t,an+1-an+2=0(t∈N*,n∈N*),记数列{an}的前n项和的最大值为f(t),
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