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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.

 Mar 9, 2019
 #1
avatar+7458 
+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 a10 is 2,646,275.

 

Notes: The general relation can be rewritten as an+1 = 5an + an-1.

 

https://oeis.org/A052918 

^ Here's a link to the sequence.

 Mar 9, 2019
 #4
avatar+14 
+1

Thanks for the two answers but for the link on the first one I am confused on how to read it.

SUS101  Mar 9, 2019
 #6
avatar+7458 
+1

You read the line that says "A052918              a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1)."

That's the index of the sequence and the recursive formula for it.

Then the list below are the members of the sequence.

MaxWong  Mar 9, 2019
 #2
avatar+7458 
+2

2.

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

.
 Mar 9, 2019
 #3
avatar+7458 
+2

3.

\(a_1 = 1\\ a_2 = \dfrac{a_1}{1+a_1} = \dfrac{1}{2}\\ a_3 = \dfrac{a_2}{1+a_2} = \dfrac{1}{3}\\ a_4 = \dfrac{a_3}{1+a_3} = \dfrac{1}{4}\\ \vdots\)

Obviously, \(a_n = \dfrac{1}{n}\quad \forall n \geq 1\).

\(\therefore a_{10} = \dfrac{1}{10}\)

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 Mar 9, 2019
 #5
avatar+7458 
+2

4.

\(\displaystyle \sum^{n}_{k=1} a_k = n(n+1)(n+2)\).

\(a_{10} = \displaystyle \sum^{10}_{k=1} a_k - \sum^9_{k=1} a_k = 10(11)(12) - 9(10)(11) = 330\)

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 Mar 9, 2019
edited by MaxWong  Mar 9, 2019

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