Question

数列an+1=an+2n+1，a1=1，求数列an通项公式(写出具体步骤！！！)

Answer 1

a1=1
a2=a1+2*1+1
a3=a2+2*2+1
……
a(n-1)=a(n-2)+2*(n-2)+1
a(n)=a(n-1)+2*(n-1)+1
将上面n个等式左右同时叠加，则a1,a2……a(n-1)都可消去
得an=2*[1+2+……+(n-1)]+1*n=n^2

Answer 2

你题目应该没有写对吧，等式两边的an可以约掉啊，看你的等式

Answer 3

an-a(n-1)=2n-1
an=(an-a(n-1))+(a(n-1)-a(n-1))+…+(a2-a1)+a1
=1+3+5+…+(2n-3)+(2n-1)=n^2

Answer 4

An=A(n-1)+2(n-1)+1
=A(n-2)+2(n-2)+1+2(n-1)+1
=...
=A(1)+2(1+2+...+n-1)+n-1
=1+2n(n-1)/2 +n-1
=1+n^2-n+n-1
=n^2