已知在等比数列｛an｝中,a2=2,a5=128,若bn=log2an,求数列｛bn｝前n项和Sn

a5=a2q^3
128=2q^3
q^3=64
q=4
an=a1q^(n-1)
=a2q^(n-2)
=2*4^(n-2)
=2*2^(2n-4)
=2^(2n-3)
bn=log2an
=log2 2^(2n-3)
=2n-3
所以bn 是以2为公差的等差数列
b1=2*1-3=-1
sn=-1+1+3+.+2n-3
=(2n-3-1)n/2
=n(n-2)
sn