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Let us consider a sequence [tex]\rm a_{n}[/tex] as follows.
[tex]\rm a_{n+1}=a_{n}(a^{2}_{n}-3a_{n}+3),\ a_{1}=10[/tex] [tex]\;[/tex]
Find [tex]a_{5}.[/tex]​

Sagot :

LammettHashidn

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

[tex]a_2 = a_1 ({a_1}^2 - 3a_1 + 3) = 10 (10^2-30+3) = 730[/tex]

Do the same to find the next few terms:

[tex]a_3 = a_2 ({a_2}^2 - 3a_2 + 3) = 730 (730^2 - 2190 + 3) = 387\,420\,490[/tex]

[tex]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[/tex]

[tex]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}[/tex]

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