Hi I was wondering if anyone could help me with the following two problems, I got 675 for the first one but it was wrong:
1. Given \(a_0 = 1\) and \(a_1 = 5,\) and the general relation
\(a_n^2 - a_{n - 1} a_{n + 1} = (-1)^n\)
for \(n \ge 1,\) find \(a_3.\)
2. Evaluate the sum \(\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\)
+129899 2. Note that we can express this in a slightly different manner :
We can write
6 6 6 6
_____________ + _____________ + ______________ + ... + ____________
[2 (1) + 1]^2 - 1 [ 2(2) + 1]^2 -1 [ 2(3) + 1]^2 - 1 [2(n) + 1]^2 -1
Note that
6 6 6 3 [ 1 1 ]
__________ = ______________ = __________ = ____ * ___ - ____
(2n +1)^2 - 1 4n^2 + 4n + 1 - 1 4 (n^2 + n) 2 [ n n + 1 ]
So we can write
(3/2) [ 1/ 1 - 1/ [ 1 + 1) + 1/ 2 - 1 / (2 + 1) + 1/3 - 1 ( 3 + 1) + 1/4 - 1/(4 + 1) +...+ 1/n - 1/n+1) ] =
(3/2) [ 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ..... + 1/n - 1/(n+1) ] =
(3/2) [ 1 - 1 / ( n + 1) ]
As n approaches some "large" number, 1/ ( n + 1) approaches 0
So.....the sum is
(3/2) [ 1 + 0 ] =
(3/2) [ 1 ] =
3/2
+2668 I will take a shot at 1.
Subsitute 1 as n.
This gives: \(5^2-1 \times a_2 = -1^1\)
Solving for the unknown, we find \(a_2 = 26\)
Subsituting 2 for n, we get: \(26^2-5a_3=-1^2\)
Solving for the unknown again, we find \(\color{brown}\boxed{a_3=135}\)
