As-time-gs-by
幼苗
共回答了17个问题采纳率:88.2% 举报
(1)由韦达定理可得:
a(n)a(n+1)=(1/3)^n
a(n+1)a(n+2)=(1/3)^(n+1)
下式÷上式得:
a(n+2)/a(n)=1/3=定值;
(2)取n=1,则a(1)a(2)=1/3,a(2)=1/6,
所以,可得:
a(2n-1)=2×(1/3)^(n-1),
a(2n)=(1/6)×(1/3)^(n-1);
(3)B(n)=a(n)+a(n+1)
当n=2k,则
B(2k)=(1/6)×(1/3)^(k-1)+2×(1/3)^k
=(5/6)×(1/3)^(k-1);
当n=2k-1,则
B(2k-1)=2×(1/3)^(k-1)+(1/6)×(1/3)^(k-1)
=(13/6)×(1/3)^(k-1);
所以
S=13/6+5/6+(13/6)×(1/3)+(5/6)×(1/3)+(13/6)×(1/3)^2+(5/6)×(1/3)^2+…+(13/6)×(1/3)^(k-1)+(5/6)×(1/3)^(k-1)+…
=(13/6+5/6)/(1-1/3)
=3÷(2/3)
=9/2
(第三不是很肯定.,请检验)
1年前
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