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I don't know if the latex will work, so sorry in advance.

 

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. 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 \($a_{10}.$\)

 

3. The first term of a sequence is 13. Starting with the second term, each term is the sum of the cubes of the digits in the previous term. For example, the second term is \($1^3 + 3^3 = 28.$\) Find the 100th term.

 

Thanks for the help in advance!

 Feb 15, 2020
 #1
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3 -  n=13;s=0;p=0;c=1;printc," - ",n;cycle: s=(n%10);p=p+s^3;n=int(n/10);if(n!=0, goto cycle,0);c++;printc," - ",p;n=p;p=0;if(c<=100, goto cycle, 0);printc," - "

 

OUTPUT =:

1  -  13
2  -  28
3  -  520
4  -  133
5  -  55
6  -  250
7  -  133
8  -  55
9  -  250
10  -  133.....................So, the 100th term = 133
 

 Feb 15, 2020
 #2
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0

1. From the recursion, a_2 = 24, and a_3 = 571.

 

2. From the recursion, a_2 = 1/2, a_3 = 2/3, a_4 = 3/4, and so on.  The tenth term a_{10} is 9/10.

 Feb 15, 2020
 #3
avatar+108776 
0

Please put only one question per post.

AND leave out the meaningless $ signs.

To me this looks like the post of a very lazy person who cannot even be bothered presenting his question properly.

 Feb 16, 2020

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