已知数列{an},a1=1,a(k+1)=(2^k) * (ak),(k≥1)

a(n+1)=(2^n)an
an=[2^(n-1)]a(n-1)
……
a2=2a1
连乘
a2a3...an=a1a2...a(n-1)[2^1×2^2×...×2^(n-1)]
an=a1[2^[n(n-1)/2]]=2^[n(n-1)/2]
bn=log2[(an)/(4^n)]=n(n-1)/2-2n=(n^2-5n)/2=n(n-5)/2
当n=2和n=3时,bn有最小项,b2=b3=-3
{cn}前n项和Sn=bn+2|S4|=n(n-5)/2+4=(n^2+3n)/2