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如题所述

(I)
a1=3
an = a1.q^(n-1)
Sn=a1+a2+...+an
b1=1
bn=b1.q^(n-1) >0
b2+a2=9
q + 3+d =9
q+d = 6 (1)
q^2=3+a2
= 6+d
= 6+ 6-q
q^2+q-12=0
(q+4)(q-3)=0
q=3
from (1)
q+d=6
3+d=6
d=3
an = 3+ 3(n-1) = 3n
bn= 3^(n-1)
(II)
Sn = a1+a2+...+an = 3n(n+1)/2
1/Sn = (2/3)[1/n(n+1) ] = (2/3) [ 1/n - 1/(n+1) ]
1/S1 +1/S2+...+1/Sn
=(2/3) [ 1 - 1/(n+1) ]
=(2/3) [n/(n+1)]
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第1个回答  2018-05-10
b2+a2=9
q+a2=9
q^2=3+a2
q^2+q-12=0
q1=3,q2=-4(bn>0,q<0舍去)。
q=3,a2=6

d=a2-a1=6-3=3
an=3+3(n-1)=3n
bn=3^(n-1)
sn=3n+3n(n-1)/2=3n(n+1)/2
1/sn=2/3[1/n-1/(n+1)]
1/s1+1/s2+...+1/sn=2/3[1/1-1/(n+1)]=2n/[3(n+1)]
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