The answer is 1830.
Let a=f(1).By induction,one may easily prove that for any n>=0,
f(4n+1)=a,
f(4n+2)=8n+1+a,
f(4n+3)=2-a,
f(4n+4)=8n+7-a.
Therefore,
f(1)+f(2)+...+f(60)
=\sum_{n=0}^{14}(f(4n+1)+f(4n+2)+f(4n+3)+f(4n+4))
=\sum_{n=0}^{14}(16n+10)
=16*14*15/2+150
=1830.??????????????????????а???
高中數(shù)學,已知數(shù)列{f（n）}滿足f（n+1）+f(n)×(-1)^n=2n-1,求此數(shù)列前60項和.
高中數(shù)學,已知數(shù)列{f（n）}滿足f（n+1）+f(n)×(-1)^n=2n-1,求此數(shù)列前60項和.
數(shù)學人氣：794 ℃時間：2020-06-24 16:23:54
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