n=1/an,b1=1/a1,a1=an/q^(n-1),1/a1=q^(n-1)/an
b(n+1)/bn=[1/a(n+1)]/(1/an)=an/a(n-1)=1/q
所以bn是以1/a1为首项,1/q为公比的等比数列
Tn=[1/a1-1/(qan)]/(1-1/q)
=[q^(n-1)/an-1/(qan)]/(1-1/q)
=(q^n/an-1/an)/(q-1)
=(q^n-1)/[an(q-1)]
n=1/an,b1=1/a1,a1=an/q^(n-1),1/a1=q^(n-1)/an
b(n+1)/bn=[1/a(n+1)]/(1/an)=an/a(n-1)=1/q
所以bn是以1/a1为首项,1/q为公比的等比数列
Tn=[1/a1-1/(qan)]/(1-1/q)
=[q^(n-1)/an-1/(qan)]/(1-1/q)
=(q^n/an-1/an)/(q-1)
=(q^n-1)/[an(q-1)]