Question

已知下列两数列｛an｝的前n项和Sn的公式，求｛an｝的通项公式(1)Sn=n³+n-1(2)Sn=x²-1

(2)Sn=n²-1

Answer 1

1、a1=S1=1
当n＞1时
an=S(n+1)-Sn=(n+1)³+(n+1)-1-(n³+n-1)=3n²+3n+2
∴an=1(n=1)
=3n²+3n+2(n＞1)
2、a1=S1=0
当n＞1时
an=S(n+1)-Sn=(n+1)²-1-(n²-1)=2n+1
∴an=0(n=1)
=2n+1(n＞1)

Answer 2

当n=1时，an=S1
当n>1时，an=Sn—Sn-1
最后检验n=1时，是否满足n>1的情况，若满足an为>1的等式，若不满足则分开写

Answer 3

1、
an=S(n)-S(n-1)
=(n^3+n-1)-[(n-1)^3+(n-1)-1]
=n^3-(n-1)^3+1
=n^3-(n^3-3n^2+3n-1)+1
=3n^2-3n+2
2、
an=S(n)-S(n-1)
=n^2-1-[(n-1)^2-1]
=n^2-(n^2-2n+1)
=2n-1

Answer 4

an=Sn-Sn-1
Sn-1=(n-1)³+n-1
an=Sn-Sn-1
=n³+n-1-(n-1)³+(n-1)-1
=3n^2-2n -2

Answer 5

n=1,an=a1,
n>1,an=sn-s(n-1)