n=1時(shí),b1/2=a1=2×1-1=1
b1=2
n≥2時(shí),
b1/2+b2/2²+...+bn/2ⁿ=an=2n-1 (1)
b1/2+b2/2²+...+b(n-1)/2^(n-1)=a(n-1)=2(n-1)-1=2n-3 (2)
(1)-(2)
bn/2ⁿ=2
bn=2^(n+1)
n=1時(shí),b1=2²=4≠2,數(shù)列{bn}的通項(xiàng)公式為
bn=2 n=1
2^(n+1) n≥2
n=1時(shí),S1=b1=2
n≥2時(shí),
Sn=b1+b2+...+bn
=2+2³+...+2^(n+1)
=2+2²+...+2^(n+1) -4
=2×[2^(n+1) -1]/(2-1) -4
=2^(n+2) -6
n=1時(shí),S1=2³-6=8-6=2,同樣滿足.
綜上,得Sn=2^(n+2) -6