Question

已知数列{an}的前n项和为Sn,且满足Sn+1-2=Sn+an+2n(n 为正数,且s2=8 求数列{an}的通项公式

Answer 1

S(n+1)-Sn=a(n+1)
所以a(n+1)-an=2n+2
所以
an-a(n-1)=2(n-1)+2
a(n-1)-a(n-2)=2(n-2)+2
……
a2-a1=2*1+2
相加
an-a1=2*[1+2+……+(n-1)]+2*(n-1)=n(n-1)+2(n-1)=n^2+n-2

令n=1
S2-2=S1+a1+2*1
S2=8,S1=a1
所以8-2=2a1+2
a1=2
所以an=n^2+n

Answer 2

移项得an+1=Sn+1-Sn=an+2n+2
所以an+1-an=2(n+1)
s2=8 => S1+a1+2=2a1+2=6 => a1=2
{an+1-an}成等差数列
an+1=sigma(ai+1-ai)+a1=(n+1)(n+2)
所以an=n(n+1)

Answer 3

已知数列{an}的前n项和为Sn,且满足Sn+1-2=Sn+an+2n;
Sn+1-Sn=an+1=an+2n+2
an+1-an=2n+2
数列{an}的通项公式an=A+n*(2n+2)
a1=A+4;
a2=A+12
s2=8,故,
s2=a1+a2=2A+18=8
A=-5;
数列{an}的通项公式an=-5+n*(2n+2)

Answer 4

Sn+1-2=Sn+an+2n可知,Sn+1-Sn=an+2n+2,又可知
an+1-an=2n+2
之后用累加,不用说了吧

Answer 5

n^2+n