数列an,a1=1,a(n+1)=an+2n+3,求an用累加法用累加法,

数列an,a1=1,a(n+1)=an+2n+3,求an用累加法用累加法,
数列an,a1=1,a(n+1)=an+2n+3,求an用累加法
用累加法,

数列an,a1=1,a(n+1)=an+2n+3,求an用累加法用累加法,
a1=1
a2-a1=5
a3-a2=7
a4-a3=9
.
a(n)-a(n-1)=2n+1
以上式子相加
a(n)=1+5+7+9+.+(2n+1)
=1+(5+2n+1)*(n-1)/2
=1+(n+3)(n-1)
=n²+2n-2

因为a1=1,an+1=an+2n+3,
所以
a2-a1=5
a3-a2=7
a4-a3=9
. . .
. . .
. . .
an-an-1=2n+1
所有项相加，发现an-a1=5+7+9+...+2n+1
=(n-1)[5+(2n+1)]/2
=(n-1)(3+n)
所以an=(n-1)(3+n)-1

A1+a2+……an+1=1+a1+2*1+3+……an+2n+3
=1+Sn+2(1+2+3+……+n)+3n
Sn+1=1+Sn+2(1+2+3+……+n)+3n
Sn+1- Sn=1+2(1+2+3+……+n)+3n

a(n+1)=a(n+1）-a(n)+a（n)-a(n-1)+...+a2-a1+a1
=2(n+n-1+n-2+...+1)+3n+1
=(n+1)^2+2(n+1)-2
所以an=n^2+2n-2

a1=1
a2=a1+1×2+3
a3=a2+2×2+3
a4=a3+32+3
省略号
a(n-1)=a(n-2)+2(n-2)+3
令an=a(n-1)+2(n-1)+3
左右两边分别累加；抵消
an=1+1×2+2×2+3×2+┈+2(n-1)+3(n-1)
=1+(2+2n-2)(n-1)/2+3(n-1)
n^2+2n-2

a(n+1)-an=2n+3
a2-a1=2+3=5
a3-a2=4+3=7
a4-a3=6+3=9
.....
an-a(n-1)=2(n-1)+3=2n+1
an-a1=(5+2n+1)(n-1)/2
an-1=n*2+2n-3
an=n*2+2n-2