设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11且（5n-8)S(n+1）-(5n+2)Sn=-20n-8求大神帮助求证：数列{an}是等差数列

Question

设数列{an}的前n项和为Sn,已知a1=1,a2=6,a3=11且（5n-8)S(n+1）-(5n+2)Sn=-20n-8求大神帮助
求证：数列{an}是等差数列

Answer 1

（5n-8)S(n+1）-(5n+2)Sn=-20n-8 (5n-8)an+(5n-8)Sn-(5n+2)Sn=-20n-8 (5n-8)an-10Sn=-20n-8 ① 令n=n-1,得 (5n-13)a(n-1)-10S(n-1)=-20n+12 ② ①-②,得 (5n-18)an-(5n-13)a(n-1)=-20 ③ 再令n=n-1,得 (5n-23)a(n-1)-(5n-18)a(n-2)=-20 ④ ③-④,得 (5n-18)an+(5n-18)a(n-2)=(5n-13)a(n-1)+(5n-23)a(n-1)=2(5n-18)a(n-1) 得 an+a(n-2)=2a(n-1) an-a(n-1)=a(n-1)-a(n-2) 所以an是等差数列!正毕!