Question

已知数列an中,a1=1,前n项和sn=(n+2/3)*an,求1/an的前n项和

Answer 1

Sn - Sn-1
= (n+2)/3 * an - (n-1+2)/3 * an-1
= (n+2)/3 * an - (n+1)/3 * an-1

所以
an = (n+2)/3 * an - (n+1)/3 * an-1
(n-1)/3 * an - (n+1)/3 * an-1 =0
整理得
(n-1) an = (n+1) an-1
an/an-1 = (n+1)/(n-1)
an/an-1 * an-1/an-2 *...* a3/a2 * a2/a1 = (n+1)/(n-1) * n/(n-2) *...* (3+1)/(3-1)* (2+1)/(2-1)
左边 = an/a1 = an
右边 = (n+1) *n / (3-1)(2-1) = (n+1)*n/2
ji即
an = (n+1)*n/2

1/an = 2/n(n+1) = 2/n - 2/(n+1)
1/an的前n项和
=1/a1 + 1/a2 + ... + 1/an
=1 + (2/2 - 2/3) + .. . + (2/n - 2/(n+1))
= 2 - 2/(n+1)
= 2n/(n+1)

Answer 2

Sn+1=(n+1+2/3)*an+1

Sn+1-Sn=a_n+1=(n+1+2/3)*a_n+1-(n+2/3)*an
得a_n+1=an=1
所以S_1/an=n\

题目本身有问题，S1不等于a1