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.
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![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.](/uploads/image/z/11682632-56-2.jpg?t=%E8%8B%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D2%2F3%2Ca2%3D2%2C3%28an%2B1-2an%2Ban-1%29%3D2%2C%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E%E6%95%B0%E5%88%97%7Ban%2B1-an%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E4%BD%BF%281%2Fa1%29%2B%281%2Fa2%29%2B%281%2Fa3%29%2B%E2%80%A6%2B%281%2Fan%29%3E5%2F2%E6%88%90%E7%AB%8B%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E6%95%B4%E6%95%B0n.)
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.
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.
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.
(1)
3[a(n+1)-2an+a(n-1)]=2
a(n+1)-2an+a(n-1)=2/3
a(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)(2/3)=(2/3)n
a(n-1)-an=(2/3)(n-1)
…………
a2-a1=(2/3)×2
累加
an-a1=(2/3)(1+2+3+...+n)
an=a1+(2/3)(2+3+...+n)
=(2/3)(1+2+...+n)
=(2/3)[n(n+1)/2]
=n(n+1)/3
1/an=3/[n(n+1)]=3[1/n-1/(n+1)]
1/a1+1/a2+...+1/an
=3[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=3[1-1/(n+1)]
=3n/(n+1)
3n/(n+1)>5/2
n>5
又n为正整数,n最小为6.
(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