Question

已知数列{an}的前n项和为Sn,求{an}的通项公式. (1)Sn=2n²-3n+k (2)Sn=3²+b 求具体步骤 谢谢！！

Answer 1

(1)an=Sn-Sn-1=2n²-3n+k-2(n-1)²+3(n-1)-k=4n+5，与推论相符，当n=1时，a1=9推出k=10
∴an=4n+5
(2)an=Sn-Sn-1=0，∴数列an为恒为0的等差数列，即an=0,推出b=-9

Answer 2

a1=s1=2-3+k=k-1
Sn=2n^2-3n+k
S(n-1)=2(n-1)^2-3(n-1)+k
=2n^2-4n+2-3n+3+k
=2n^2-7n+5+k

an=sn-s(n-1)
=2n^2-3n+k-(2n^2-7n+5+k)
=2n^2-3n+k-2n^2+7n-5-k
=-4n-5
an=4n-5(n>=2)

a1=a1=9+b
Sn=3²+b=9+b
s(n-1)=9+b
an=sn-s(n-1)=0

an=0(n>=2)