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Let \(\{a_n\}\) be an arithmetic sequence with \(a_3=7\) and \(a_5+a_7=26\). For any positive number \(n\), define

 

\(b_n = \frac{1}{a_n^2-1}\).

 

Suppose \(b_1+b_2+b_3...+b_{2000}=p/q,\) where \(p\) and \(q\) are relatively prime positive integers. What is \(p-q?\)

 

 

 

 

I found that the first seven of the arithmetic sequence are

\(3,5,7,9,11,13,15\)

From that, I figured that \(a_n=1+2n\), thus \(b_n=\frac{1}{(1+2n)^2-1}=\frac{1}{4n^2+4n}\).

That's where I got stuck. I can't think of a way to sum the terms. Can somebody help me? Thanks in advance!

 Aug 2, 2022
 #1
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Nice work figuring out the arithmetic series  !!!

 

b1999 =  2(1999) + 1  =  3999

b 2000 =  2(2000)   +1  =  4001

 

bn  =      1  / [ (a)^2  - 1] 

 

Note that we can write this as partal fractions as follows :

 

bn  =   1 / [ ( an - 1) (an + 1)]     =   

 

1/ [ 2 (an -1) ]  - 1/ [2(an + 1) ]  =

 

 (1/2) [ 1/(an - 1) - 1/(an + 1)]

 

So  we can write

 

(1/2)  [  ( 1/2 - 1/4 ) +  (1/4  - 1/6) + ( 1/6 - 1/8) + ( 1/8 - 1/10)  +  ....... + ( 1/3998 - 1/4000) + (1/ 4000 - 1/4002 )

 

 

Notice that most of the terms  "cancel"   and we are left with

 

(1/2) ( 1/2  - 1/4002)  =

 

(1/2)  [ 4002 - 2 ]  / ( 8004)    =

 

(1/2) (4000)  / 8004  =

 

2000 / 8004

 

500/ 2001    =  p/q

 

p - q    =   500 - 2001   =  -1501

 

 

cool cool cool

 Aug 2, 2022
 #2
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Thank you!

 Aug 2, 2022

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