设数列an满足a1=a2=1,a3=2,且对正整数n都有an·an+1·an+2·an+3=an+an+1+an+2+an+3,
问题描述:
设数列an满足a1=a2=1,a3=2,且对正整数n都有an·an+1·an+2·an+3=an+an+1+an+2+an+3,
,求a1+a2+a3+···+a100的值
答
a1×a2×a3×a4=a1+a2+a3+a4
1×1×2×a4=1+1+2+a4
a4=4
a2×a3×a4×a5=a2+a3+a4+a5
1×2×4×a5=1+2+4+a5
7a5=7
a5=1=a1
a3×a4×a5×a6=a3+a4+a5+a6
2×4×1×a6=2+4+1+a6
7a6=7
a6=1=a2
a4×a5×a6×a7=a4+a5+a6+a7
4×1×1×a7=4+1+1+a7
3a7=6
a7=2=a3
…………
数列{an}是周期为4的周期数列,从第1项开始,按1,1,2,4循环,每4项循环一次.
100÷4=25,正好循环25次.
a1+a2+a3+...+a100=25×(1+1+2+4)=25×8=200