已知数列{an}的前n项和为Sn,求{an}的通项公式。(1)Sn=2n^2-3n;(2)Sn=3^n+6

2025-05-09 16:12:25
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回答1:

(1) an=Sn-S(n-1) =2n^2-3n-[2(n-1)^2-3(n-1)] =2n^2-3n-(2n^2-4n+2-3n+3 =4n-5(2) an=Sn-S(n-1) =3^n+6-[3^(n-1)+6] =3^n+6-3^(n-1)-6 =3^n-3^(n-1) =3*3^(n-1)-3^(n-1) =2*3^(n-1)