+158 I don't know if the latex will work, so sorry in advance.
1. Given $a0=1$ and $a1=5,$, and the general relation a2n−an−1an+1=(−1)n for $n≥1,$ find $a3.$
2. A sequence ${an}$ satisfies $a1=1$ and an=an−11+an−1 for all $n≥2.$ Find $a10.$
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 $13+33=28.$ Find the 100th term.
Thanks for the help in advance!
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
