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# Telescoping Problems

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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 $$\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots$$

Mar 1, 2022

#1
+1

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   Mar 2, 2022
#2
+1

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 $$\color{brown}\boxed{a_3=135}$$

Mar 2, 2022