已知等比数列an满足an+1十an=9x2的n一1次幂求an的通项公式及前n项和Sn

Question

已知等比数列an满足an+1十an=9x2的n一1次幂求an的通项公式及前n项和Sn

Answer 1

由题意知,an为等比数列,设公比为q.a(n+1) + an = 9*2^(n-1)
于是 a1+a2=9 ,a2=a1q => a1(1+q)=9------------------(1)
a2+a3=18 ,a2=a1q a3=a1*q^2 => a1q(1+q)=18---------------(2)
联立(1)(2)消去a1 ,又因为q+1不等于0 ,解得q=2,a1=3
an=3*2^(n-1)
Sn=3(2^n-1)
祝愉快