等差数列{an}的前n项和为Sn,a1=1,S10=100,若有数列{bn},满足an=log2bn,则b1+b2+b3+b4+b5=?

Question

等差数列{an}的前n项和为Sn,a1=1,S10=100,若有数列{bn},满足an=log2bn,则b1+b2+b3+b4+b5=?

Answer 1

设公差为d,则
s10=(a1+a10)*10/2=5*(a1+a1+9d)=10+45d
即 10+45d=100 d=2
由an=log2bn 知 bn=2^an
则：
b1+b2+b3+b4+b5=2^a1+2^a2+2^a3+2^a4+2^a5=2^1+2^3+2^5+2^7+2^9=682