已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012
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已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012已知数列{an}满足a1=1
已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012
已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012
已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012
a1=1 a2=3
a3=a2-a1=3-1=2 a4=a3-a2=2-3=-1
a5=a4-a3=-1-2=-3 a6=a5-a4=-3-(-1)=-2
a7=a6-a5=-2-(-3)=1 a8=a7-a6=1-(-2)=3
…………
数列从第1项开始,按1、3、2、-1、-3、-2循环,每6项循环一次,每个循环的和:
1+3+2+(-1)+(-3)+(-2)=0
2012=6×335+2,循环335次后剩余两项.
S2012=0+a2011+a2012=a1+a2=1+3=4
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