Question

高一数学，第二问，求详细解释

Answer 1

解：
(1)
a(n+1)-an=4，为定值，数列{an}是以4为公差的等差数列
a1+a4=14
2a1+3d=14
a1=(14-3d)/2=(14-3×4)/2=1
an=a1+(n-1)d=1+4(n-1)=4n-3
数列{an}的通项公式为an=4n-3
(2)
Sn=(a1+an)n/2=(1+4n-3)n/2=n(2n-1)
b1=S1/(1+k)=a1/(1+k)=1/(k+1)
bn=Sn/(n+k)=n(2n-1)/(n+k)
b(n+1)-bn
=(n+1)[2(n+1)-1]/(n+1+k) -n(2n-1)/(n+k)
=(n+1)(2n+1)/(n+k+1) -n(2n-1)/(n+k)
=[(n+1)(2n+1)(n+k)-n(2n-1)(n+k+1)]/[(n+k+1)(n+k)]
=[2n²+(4k+2)n+k]/[(n+k+1)(n+k)]
{bn}是等差数列，[2n²+(4k+2)n+k]/[(n+k+1)(n+k)]与n的取值无关，令为d'
[2n²+(4k+2)n+k]/[(n+k+1)(n+k)]=d'
整理，得
(d'-2)n²+(2kd'+d'-4k-2)n+k(d'k+d'-1)=0
要等式左边与n的取值无关
d'-2=0，2kd'+d'-4k-2=0，k(d'k+d'-1)=0
解得d'=2，k=0或k=-½
k=0时，b1=1，bn=2n-1
1/[bnb(n+1)]=1/[(2n-1)(2n+1)]=½[1/(2n-1) -1/(2n+1)]
Tn=1/(b1b2) +1/(b2b3)+...+1/[bnb(n+1)]
=½[1- 1/3 +1/3- 1/5+...+1/(2n-1) -1/(2n+1)]
=½[1- 1/(2n+1)]
=n/(2n+1)
k=-½时，b1=2，bn=2n
1/[bnb(n+1)]=1/[2n·2(n+1)]=¼[1/n -1/(n+1)]
Tn=1/(b1b2)+1/(b2b3)+...+1/[bnb(n+1)]
=¼[1- 1/2 +1/2 -1/3+...+1/n -1/(n+1)]
=¼[1- 1/(n+1)]
=n/[4(n+1)]
综上，得：Tn=n/(2n+1)或Tn=n/[4(n+1)]
注意第二问有两解。