若数列{an}满足:a1=2/3,a2=2,3(an+1-2an+an-1)=2,(1)证明数列{an+1-an}是等差数列;(2)求使(1/a1)+(1/a2)+(1/a3)+…+(1/an)>5/2成立的最小正整数n.
问题描述:
若数列{an}满足:a1=2/3,a2=2,3(an+1-2an+an-1)=2,
(1)证明数列{an+1-an}是等差数列;
(2)求使(1/a1)+(1/a2)+(1/a3)+…+(1/an)>5/2成立的最小正整数n.
答
(1)由 3(an+1-2an+an-1)=2 得 3(an-2an-1+an-2)=2 3(an-1-2an-2+an-3)=2 ......3(a3-2a2+a1)=2(竖着写看的比较清楚)等式左边全加起来=an+1-an-a2+a1
等式右边全加起来=2(n-1) 即an+1-an-a2+a1=2(n-1) 即an+1-an=2(n-1)+4/3=2n-2/3
{an+1-an}-{an-an-1}=2 所以{an+1-an}为等差数列
答
(1)3[a(n+1)-2an+a(n-1)]=2a(n+1)-2an+a(n-1)=2/3a(n+1)-an-[an-a(n-1)]=2/3,为定值.a2-a1=2-2/3=4/3数列{a(n+1)-an}是以4/3为首项,2/3为公差的等差数列.(2)a(n+1)-an=4/3+(n-1)(2/3)=(2/3)(n+1)an-a(n-1)=4/3+(n-1)...