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!
+128460 Nice work figuring out the arithmetic series !!!
b1999 = 2(1999) + 1 = 3999
b 2000 = 2(2000) +1 = 4001
bn = 1 / [ (an )^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
