Question

已知Sn为数列{an}的前n项和,a1=a,an+1=Sn+3^n求大神帮助

Answer 1

1.an+1=Sn+3^n an=Sn-1 + 3^n-1 相减得an+1 - an =Sn - Sn-1 + 2*3^n-1 即an+1=2an + 2*3^n-1 an+1 - 2*3^n =2(an - 2*3^n-1) {an - 2*3^n-1}是q=2的 等比数列 所以an - 2*3^n-1=(a-2)*2^n-1 ,an=(a-2)*2^n-1 + 2*3^n-1 Sn=an+1 - 3^n =(a-2)*2^n + 2*3^n - 3^n=(a-2)*2^n + 3^n 故bn=(a-2)*2^n 2.an+1 >=an (a-2)*2^n + 2*3^n >=(a-2)*2^n-1 + 2*3^n-1 (a-2)*2^n-1 >= -4*3^n-1 因为恒成立 所以a - 2>=0 即 a>=2.