an>0,则首项a1>0,公比q>0a2a4+2a3a5+a4a6=25a1²q^4+2a1²q^6+a1²q8=25a1²q^4(1+2q²+q⁴)=25a1²q^4(q²+1)²=25a1q²(q²+1)=5a1q²+a1q⁴=5a3+a5=5...