设k是给定的正整数，对于满足条件a1-a2k+1=2的所有无穷等差数列{an}，ak+1+ak+2+…+a2k+1的最大值．

Question

设k是给定的正整数，对于满足条件a₁-a

2

k+1

=2的所有无穷等差数列{a_n}，a_k+1+a_k+2+…+a_2k+1的最大值___．

Answer 1

设等差数列{a_n}的公差为d，
∵a_k+1=a₁+kd，∴kd=a_k+1-a₁，
又a₁=a

2

k+1

+2，
∴a_k+1+a_k+2+…+a_2k+1=（k+1）a_k+1+(k(k+1))/(2)d=（k+1）(ak+1+(1)/(2)kd)=（k+1）[ak+1+(1)/(2)(ak+1-a1)]=（k+1）((3)/(2)ak+1-(1)/(2)

a

2 k+1

-1)=(k+1)/(2)[-(ak+1-(3)/(2))2+(1)/(4)]，
当且仅当a_k+1=(3)/(2)时，a_k+1+a_k+2+…+a_2k+1的最大值为(k+1)/(8)．
故答案为：(k+1)/(8)．