设{an}的前n项和为S,已知a1=1,a2=6,a3=11,(5n-8)Sn+1-(5n+2)Sn= -20n-8.n

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设{an}的前n项和为S,已知a1=1,a2=6,a3=11,(5n-8)Sn+1-(5n+2)Sn= -20n-8.n=1,2,3...求证{an}为等差数列.
很难啊

(5n-8)Sn+1-(5n+2)Sn= -20n-8
(5n-13)Sn-(5n-3)Sn-1= -20n+12
相减
因为[sn+1-sn=an+1]
所以(5n-8)an+1+5sn-(5n+2)an-5sn-1=-20
[5n-8]an+1-(5n-3)an=-20
[5n-13]an-(5n-8)an-1=-20
相减
(5n-8)(an+1-an)-(5n-8)(an-an-1)=0
(5n-8)(an+1-2an+an-1)=0
an+1-2an+an-1=0
an-an-1=5
{an}为等差数列.
这下对了,还有问题么?

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