已知数列{an}的前n项和为Sn=n平方+2n求通项公式an

2025-05-09 02:56:38
推荐回答(2个)
回答1:

Sn=n平方+2n
S(n-1)=(n-1)²+2(n-1)
an=Sn-S(n-1)
=[n²-(n-1)²]+[2n-2(n-1)]
=(n+n-1)(n-n+1)+2(n-n+1)
=2n-1+2
=2n+1

回答2:

解:∵Sn=n²+2n
∴S(n-1)=(n-1)²+2(n-1)
∴an=Sn-S(n-1)
=n²+2n-[(n-1)²+2(n-1)]
=2n+1