Question

设数列{an}的首项a1=1，前n项和为S，且nan-Sn=2n(n-1)，求a2的值及数列{an}的通项公式an

设数列{an}的首项a1=1，前n项和为S，且nan-Sn=2n(n-1)，求a2的值及数列{an}的通项公式an

Answer 1

2a2-S2=2*2*1
S2=a1+a2=1+a2
所以2a2-1-a2=4
a2=5

nan-Sn=2n(n-1) (1)
(n-1)a(n-1)-S(n-1)=2(n-1)(n-2) (2)
(2)-(1)
Sn-S(n-1)=an
所以(n-1)a(n-1)-nan+an=2(n-1)(n-2-n)
(n-1)a(n-1)-(n-1)an=-4(n-1)
所以an-a(n-1)=4
等差，d=4
a1=1
所以an=4n-3

Answer 2

2a2-s2=4
2a2-1-a2=4
a2=5
由nan-Sn=2n(n-1)得sn=nan-2n(n-1)
an=sn-sn-1=nan-(n-1)an-1-4n+4
(1-n)an=-(n-1)an-1-4n+4
(n-1)an=(n-1)an-1+4n-4
an=an-1=4
an=a1+(n-1)d=1+4(n-1)=4n-3