Question

数列{an}满足a(n+1)+an=4n-3(n∈N*)，若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1

求过程。谢谢

Answer 1

因为a(n+1)+an=4n-3

所以S(2n+1)=a1+a2+a3+a4+a5+...+a(2n)+a(2n+1)
=a1+(a2+a3)+(a4+a5)+...+[a(2n)+a(2n+1)]
=2+(4*2-3)+(4*4-3)+...+(4*2n-3)
=2+[4*2*(1+2+...+n)]-3n
=2+4n(n+1)-3n
=4n^2+n+2