1) Given that \( a_0 = 1\) and \( a_1 = 5 \), and the general relation \(a_n^2 - a_{n - 1} a_{n + 1} = (-1)^n\)

for \( n \ge 1,\) find \(a_{10}.\)

2)Evaluate the sum. \($\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots$.\)

3)\(A sequence $\{a_n\}$ satisfies $a_1 = 1$ and \[a_n = \frac{a_{n - 1}}{1 + a_{n - 1}}\]for all $n \ge 2.$ Find $a_{10}.$ \)

4)The sum of the first n terms of a certain sequence is n(n+1)(n+2) Find the tenth term of the sequence.

SUS101 Mar 9, 2019

#1**+1 **

1.

(Wow, this is some really advanced homework)

The generated sequence is 1,5,26,135,701,..., which is OEIS A052918 (Took me a while to find it). And a_{10} is **2,646,275**.

Notes: The general relation can be rewritten as a_{n+1} = 5a_{n} + a_{n-1}.

https://oeis.org/A052918

^ Here's a link to the sequence.

MaxWong Mar 9, 2019

#2**+2 **

2.

\(\dfrac{6}{3^2-1} + \dfrac{6}{5^2-1} +...\\ = \displaystyle 6\sum^{\infty}_{n=0} \dfrac{1}{(2n+3)^2 - 1}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \dfrac{1}{(n+1)(n+2)}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=0}\dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=1}\dfrac{1}{n+1}\right)\\ = \dfrac{3}{2}\left(\dfrac{1}{0+1}\right)\\ =\dfrac{3}{2} \)

MaxWong Mar 9, 2019