{An},A1=1 A [n+1]=An+2n+1 求{an}通项式?A [n+1]=An+2n+1 是 A [n+1]=An+2n-1 打错了

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{An},A1=1A[n+1]=An+2n+1求{an}通项式?A[n+1]=An+2n+1是A[n+1]=An+2n-1打错了{An},A1=1A[n+1]=An+2n+1求{an}通项式?A[n+

{An},A1=1 A [n+1]=An+2n+1 求{an}通项式?A [n+1]=An+2n+1 是 A [n+1]=An+2n-1 打错了
{An},A1=1 A [n+1]=An+2n+1 求{an}通项式?
A [n+1]=An+2n+1 是 A [n+1]=An+2n-1 打错了

{An},A1=1 A [n+1]=An+2n+1 求{an}通项式?A [n+1]=An+2n+1 是 A [n+1]=An+2n-1 打错了
那么根据题中条件有：
方法一：a1＝1
a2＝a1 + 2×1 - 1
a3＝a2 + 2×2 - 1
……
an＝a(n-1) + 2×(n-1)-1 . （n≥2）
上面各式两端分别相加得：
S(n)＝1 + S(n-1) + 2×[1+2+3…+(n-1)] - (n-1)
则 an＝S(n)-S(n-1)＝1 + 2×[1+2+3…+(n-1)] - (n-1)＝ n^2 - 2n + 2 .
即 an＝n^2 - 2n + 2 .
经检验,a1＝1 也满足此通项式.
方法二：an+1-an=2n-1
an-an-1=2n-3
.
.
.
a3-a2=3
a2-a1=1
将上面式子左右分别相加,得
an+1-a1=n^2
an+1=n^2+1
则通项公式an=(n-1)^2+1=n^2 - 2n + 2

利用累加法a(n+1)=an+2n-1
故a(n+1)-an=2n-1
an -a(n-1)=2（n-1）-1
. .
. .
. .
a2 -a1 =1
加得a(n+1)-a1 =n(n+1)-n
故an=（n-1）^2+a1
通项公式为an=（n-1）^2+1

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