Question

若数列【an】满足a1=2/3 a2=2 3*（An+1-2An+An-1）=2 证明数列【An+1-An】为等差数列 (2)求...

若数列【an】满足a1=2/3 a2=2 3*（An+1-2An+An-1）=2 证明数列【An+1-An】为等差数列 (2)求1/a1+ .....+1/an和Sn

Answer 1

(1)
3[a(n+1)-2an+a(n-1)] =2
3[a(n+1) - an] - 3[an - a(n-1)] =2
[a(n+1) - an] - [an - a(n-1)] =2/3
=>{a(n+1) -an} 为等差数列
(2)
[a(n+1) - an] - [an - a(n-1)] =2/3
[a(n+1) - an] - [a2 - a1] =2(n-1)/3
[a(n+1) - an] - 4/3 =2(n-1)/3
a(n+1) - an = 2(n+1)/3
an - a(n-1) = 2n/3
an -a1 = (4/3+ 2n/3)(n-1)/2
= (n+2)(n-1)/3
an = (n+2)(n-1)/3 +2/3
= (n^2+n)/3
=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/(n+1)]

an = n(n+1)/3
= (1/9)[n(n+1)(n+2) -(n-1)n(n+1)]
Sn = a1+a2+...+an
=(1/9)n(n+1)(n+2)