+0

# Just a few questions

-1
307
3

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

### 3+0 Answers

#1
0

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
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
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