Question

Let us consider a sequence }" align="absmiddle" class="latex-formula"> as follows.
$\rm a_{n+1}=a_{n}(a^{2}_{n}-3a_{n}+3),\ a_{1}=10$ $\;$
Find $a_{5}.$​

Answer 1

Use the given recursive rule to find the second term, a₂ :

$a_2 = a_1 ({a_1}^2 - 3a_1 + 3) = 10 (10^2-30+3) = 730$

Do the same to find the next few terms:

$a_3 = a_2 ({a_2}^2 - 3a_2 + 3) = 730 (730^2 - 2190 + 3) = 387\,420\,490$

$a_4 = a_3 ({a_3}^2 - 3a_3 + 3) = 387\,420\,490 (387\,420\,490^2 - 1\,162\,261\,470 + 3)\\ = 58\,149\,737\,003\,040\,059\,690\,390\,170$

$a_5 = a_4 ({a_4}^2 - 3a_4 + 4) = \cdots \\ = 196\,627\\{}\,\,\,\,\,\,050\,475\,552\,913\,618\,075\,908\,526\\{}\,\,\,\,\,\,912\,116\,283\,103\,450\,944\,214\,766\\{}\,\,\,\,\,\,927\,315\,415\,537\,966\,391\,196\,810 \\ \approx 1.96 \times 10^{27}$

Answer 2

Hi there,

$a_{5}=2850$

Please check the attached image for answer.

Hope it helps ...