a2=2,a5=16→公比q=2设a1a2+a2a3+…+ana(n+1)=Sn则q²Sn=4Sn=a2a3+a3a4+…+a(n+1)a(n+2)上式减下式得：-3Sn=a1a2-a(n+1)a(n+2)=2(a1²-a(n+1)²根据公比求得a1=1，a(n+1)=2^n于是Sn=2[1-2^2n]/(-3)=2/3×(4^n-1)
2*4^n
