In the arithmetic sequence an, the sum of the first n terms is Sn, BN = 1 / Sn, B4 = 1 / 10, s6-s3 = 15. Find the general term of BN

In the arithmetic sequence an, the sum of the first n terms is Sn, BN = 1 / Sn, B4 = 1 / 10, s6-s3 = 15. Find the general term of BN

B4=1/10
S4=10 4A1+6d=10
S6-S3=15 6A1+15d-3A1-3d=15
A1=1 d=1
An=n
Sn=n(n+1)/2
Bn=2/n(n+1)=2[1/n-1/(n+1)]
If the sum of the first n items of BN is also required, the same items will be eliminated

The general formula of sequence an is an = 2n + 1, BN = 1/

a1+a2+a3...+an=n(n+2)bn=1/(n(n+2))=(1/2)(n+2-n)/(n(n+2))=(1/2)(1/n-1/(n+2))Sbn=(1/2)(1/1-1/3+1/2-1/4+1/3-1/5+.+1/n-1/(n+2))=(1/2)(1+1/2-1/(n+1)-1/(n+2))=(3n^2+5n)/(4n^2+12n+8)

Given an = 1 / (3 ^ n), let BNN / an find the first n terms and Sn of sequence BN

BN = n / an = n * 3 ^ NSN = 1 * 3 + 2 * 3 ^ 2 +... + n * 3 ^ n3sn = 1 * 3 ^ 2 + 2 * 3 ^ 3 +... + n * 3 ^ (n + 1) subtract - 2Sn = (3 + 3 ^ 2 + 3 ^ 3 +... + 3 ^ n) - n * 3 ^ (n + 1) = 3 * (3 ^ n-1) / (3-1) - n * 3 ^ n * 3 = 3 / 2 * 3 ^ n-3 / 2-3n * 3 ^ n = (3 / 2-3n) * 3 ^ n-3 / 2, so Sn = (3N / 2-3 / 4) * 3 ^ n + 3 / 4

The known sequence satisfies: A1 = 1, a (n + 1) = an + 1, n is odd; 2An, n is even, let BN = a2n-1,
(I) find B2, B3, and prove that B (n + 1) = 2bn + 2;
(II) ① prove that the sequence {BN + 2} is equal ratio sequence;
② If a 2K, a (2k + 1) and 9 + a (2k + 2) form an equal ratio sequence, find the value of positive integer K

(I)
a(n+1)=an +1 ,n is odd
=2an ,n is even
bn=a(2n) -1
a1=1
a2 =a1+1 =2
if n is odd,
a(n+1) = an +1
= 2a(n-1) +1
a(n+1) +1 = 2[ (a(n-1) +1 ]
a(n+1) +1 = 2^[( n-1)/2 ].(a2 +1 )
= 3.2^[( n-1)/2 ]
a(n+1) = -1+3.2^[( n-1)/2 ]
n is odd => n= 2m-1
a(2m) =-1+3.2^(m-1)
bn = a(2n) -1
=-2+3.2^(n-1)
b2 = -2+3.2 = 4
b3= -2+3.4= 10
b(n+1) =-2+3.2^n
=2(-2+3.2^(n-1)) +2
=2bn +2
(II)
(1)
bn+2 =3.2^(n-1)
{BN + 2} is an equal ratio sequence, q = 3
(2)
if n is even,
a(n+1)=2an
= 2(a(n-1)+1)
a(n+1)+2 =2(a(n-1)+2)
= 2^(n/2).(a1+2)
=3.2^(n/2)
a(n+1) =-2+3.2^(n/2)
n is even ,n=2k
a(2k+1) =-2+3.2^k
A (2k), a (2k + 1), 9 + a (2k + 2) are equal ratio sequence
a(2k).[9+a(2k+2)]= [a(2k+1)]^2
[-1+3.2^(k-1) ].(8+3.2^k ) =(-2+3.2^k)^2
-8-3.2^k+12.2^k +9.2^(2k-1) = 4 -12.2^k + 9.2^(2k)
(9/2).2^(2k) -21.2^k +12=0
9.2^(2k) -42.2^k +24=0
(2^k -4)(9.2^k-6)=0
2^k=4
k=2