Question

数列｛an｝满足a1=1/2,当n〉﹦2时,2an=[a(n-1)]^2+2a(n-1)

bn=1/(2+an),数列｛bn｝的前n项和sn,前n项积tn 求证,当n属于正整数时 2^n+1 *tn +sn为定值

Answer 1

1. 求Sn：

   由2an=[a(n-1)]^2+2a(n-1)提取公因式得

   2an=(a(n-1)+2)*a(n-1)取倒数得

   1/an=2/[(a(n-1)+2)*a(n-1)]=[(a(n-1)+2)-a(n-1)]/[(a(n-1)+2)*a(n-1)]=1/a(n-1)-1/[a(n-1)+2]移项得

   1/[a(n-1)+2]=1/a(n-1)-1/an即

   b(n-1)=1/[a(n-1)+2]=1/a(n-1)-1/an即

   bn=1/an-1/a(n+1)

   得前n项和为

   Sn=b1+b2+...+bn=1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1)=1/a1-1/a(n+1)=2-1/a(n+1);

2. 求Tn：

   由2an=[a(n-1)]^2+2a(n-1)提取公因式得

   2an=(a(n-1)+2)*a(n-1)两边同除因子2*(a(n-1)+2)*an得

   1/(a(n-1)+2)=a(n-1)/(2*an)即

   b(n-1)=a(n-1)/(2*an)即

   bn=an/(2*a(n+1))

   得前n项积为

   Tn=b1*b2*...*bn=[a1/(2*a2)]*[a2/(2*a3)]*...*[an/(2*a(n+1))]=a1/[(2^n)*a(n+1)]

   =1/[(2^(n+1))*a(n+1)]

3. 将Sn和Tn结果带入2^n+1 *Tn +Sn中得

   2^(n+1)*Tn +Sn=2^(n+1)*1/[(2^(n+1))*a(n+1)]+2-1/a(n+1)=1/a(n+1)+2-1/a(n+1)=2

4. 证毕。