已知数列{an}的前n项和为Sn=3n^2+8n,则它的通项公式An等于A6n+5B6n-5C6n-1D6n+11详细解答理解的范围角度讲

Question

已知数列{an}的前n项和为Sn=3n^2+8n,则它的通项公式An等于 A 6n+5 B 6n-5 C 6n-1 D 6n+11 详细解答 理解的范围角度讲

Answer 1

n＝1时,a1＝S1＝11
n≥2时
an＝Sn－S(n-1)
＝3n²＋8n－3(n-1)²－8(n-1)
＝6n－3＋8
＝6n＋5
n＝1时,a1＝11,满足通项
所以,an＝6n＋5
选 A. 6n＋5