已知等差数列an+1+an=4n-3求a1

问题描述:

已知等差数列an+1+an=4n-3求a1
①当a1=2时,求Sn

a(n+1)=a1+nd
an=a1+(n-1)d
a(n+1)+an=2a1+(2n-1)d=4n-3
整理,得
2dn+2a1-d=4n-3
2d=4
2a1-d=-3
解得
d=2 a1=-1/2
a1=-1/2