要详解已知等比数列,(an)为递增数列,且a5方=A10,2(an+a(n+2))=5a1,求通项公式
问题描述:
要详解
已知等比数列,(an)为递增数列,且a5方=A10,2(an+a(n+2))=5a1,求通项公式
答
an=a1q^(n-1)(a5)^2 = a10(a1)^2.q^8 = a1q^9a1= q2[an+a(n+2)]=5a1an+a(n+2)=5q/2a(n+2) -5q/4= -(an -5q/4)[a(n+2) -5q/4] /(an -5q/4) = -1when n is odd(an -5q/4)/(a1 -5q/4) = (-1)^[(n-1)/2]an -5q/4 = -(q/4...