懒猫鼠
幼苗
共回答了21个问题采纳率:100% 举报
(I)a2=a1+(-1)1=0,
a3=a2+31=3.
a4=a3+(-1)2=4,
a5=a4+32=13,
所以,a3=3,a5=13.
(II)a2k+1=a2k+3k
=a2k-1+(-1)k+3k,
所以a2k+1-a2k-1=3k+(-1)k,
同理a2k-1-a2k-3=3k-1+(-1)k-1,
a3-a1=3+(-1).
所以(a2k+1-a2k-1)+(a2k-1-a2k-3)++(a3-a1)
=(3k+3k-1++3)+[(-1)k+(-1)k-1++(-1)],
由此得a2k+1-a1=[3/2](3k-1)+[1/2][(-1)k-1],
于是a2k+1=
3k+1
2+
1
2(-1)k-1.
a2k=a2k-1+(-1)k=
3k
2+
1
2(-1)k-1-1+(-1)k=
3k
2+
1
2(-1)k=1.
{an}的通项公式为:
当n为奇数时,an=
3
n+1
2
2+(-1)
n-1
2×
1
2-1;
当n为偶数时,an=
3
n
2
2+(-1)
n
2×
1
2-1.
1年前
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