Question

高中数列计算题。求详细解题过程，多谢

Answer 1

Answer 2

(4)
(I)
an = Sn/n + 2n -2

n=2
a2 = 6/2 + 4 -2
=5
a1= S2 -a2 = 6-5=1

an = Sn/n + 2n -2
Sn = n(an-2n+2)

for n>=2

an = Sn -S(n-1)
= n(an-2n+2) - (n-1)[a(n-1)-2(n-1)+2]
=nan -(n-1)a(n-1) -2n^2 +2n + (2n-4)(n-1)

=nan -(n-1)a(n-1) -2n^2 +2n + 2n^2-6n+4
=nan -(n-1)a(n-1) -4n+4
(n-1)an =(n-1)a(n-1) +4(n-1)
an = a(n-1) +4
an-a(n-1) =4
an -a1= 4(n-1)
an =4n-3

(II)
Sn = a1+a2+...+an
= (2n-1)n
S1 = a1= 1
S2 = 6
S3 = a3+S2 = 3(6-1) + 6=21

for n>=4
1/Sn = 1/[n(2n-1)]
=(1/2) { 1/[n(n-1/2)] }
<(1/2) { 1/[n(n-1)] }
= (1/2) ( 1/(n-1) -1/n)
1/S4+1/S5 +...+1/Sn < (1/2)( 1/3- 1/n)
< 1/6

1/S1+1/S2+...+1/Sn
< 1+ 1/6 +1/21+1/6
=(42+7+2+7)/42
=58/42
<5/3