Question

求一道数学题的详细过程!急!在等比数列{an}中,各项都是正数,且满足a3-a2=2a1,a4=4.(1)求数列{an}

Answer 1

(1)a3-a2=a1q^2-a1q=2a1,则q^2-q-2=0,(q+1)(q-2)=0.由题意知,q+1>0,则q=2.

a4=a1q^3=8a1=4,则a1=1/2.数列{an}的通项公式为an=(1/2)*2^(n-1)=2^(n-2),n为正整数.

(2)8an=8*2^(n-2)=2^(n+1),则bn=log2[2^(n+1)]=n+1,是首项为2、公差为1的等差数列.

b1+b2+…+b10=10*(2+10+1)/2=65.