A recurrence sequence I can't solve (A level) - Any hint is appreciated.












0












$begingroup$


A sequence is defined by
$$begin{aligned} u_1 &= 3 \
u_{n+1} &= 2 — 4/u_n
end{aligned}$$



Find the exact values of



(a) $u_2, u_3 ;text{and}; u_4$



(b) $u_{61}$



(c) $sum_{i=1}^{99} u_i$



How to solve (b) and (c) without solving the preceding terms leading up to $u_{60}$ individually ?










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$endgroup$












  • $begingroup$
    What have you tried so far? Have you done a)?
    $endgroup$
    – Mindlack
    Jan 6 at 17:51










  • $begingroup$
    Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
    $endgroup$
    – Antonio
    Jan 6 at 17:52












  • $begingroup$
    No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
    $endgroup$
    – Mindlack
    Jan 6 at 17:54
















0












$begingroup$


A sequence is defined by
$$begin{aligned} u_1 &= 3 \
u_{n+1} &= 2 — 4/u_n
end{aligned}$$



Find the exact values of



(a) $u_2, u_3 ;text{and}; u_4$



(b) $u_{61}$



(c) $sum_{i=1}^{99} u_i$



How to solve (b) and (c) without solving the preceding terms leading up to $u_{60}$ individually ?










share|cite|improve this question









$endgroup$












  • $begingroup$
    What have you tried so far? Have you done a)?
    $endgroup$
    – Mindlack
    Jan 6 at 17:51










  • $begingroup$
    Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
    $endgroup$
    – Antonio
    Jan 6 at 17:52












  • $begingroup$
    No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
    $endgroup$
    – Mindlack
    Jan 6 at 17:54














0












0








0





$begingroup$


A sequence is defined by
$$begin{aligned} u_1 &= 3 \
u_{n+1} &= 2 — 4/u_n
end{aligned}$$



Find the exact values of



(a) $u_2, u_3 ;text{and}; u_4$



(b) $u_{61}$



(c) $sum_{i=1}^{99} u_i$



How to solve (b) and (c) without solving the preceding terms leading up to $u_{60}$ individually ?










share|cite|improve this question









$endgroup$




A sequence is defined by
$$begin{aligned} u_1 &= 3 \
u_{n+1} &= 2 — 4/u_n
end{aligned}$$



Find the exact values of



(a) $u_2, u_3 ;text{and}; u_4$



(b) $u_{61}$



(c) $sum_{i=1}^{99} u_i$



How to solve (b) and (c) without solving the preceding terms leading up to $u_{60}$ individually ?







sequences-and-series






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 17:46









AntonioAntonio

11




11












  • $begingroup$
    What have you tried so far? Have you done a)?
    $endgroup$
    – Mindlack
    Jan 6 at 17:51










  • $begingroup$
    Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
    $endgroup$
    – Antonio
    Jan 6 at 17:52












  • $begingroup$
    No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
    $endgroup$
    – Mindlack
    Jan 6 at 17:54


















  • $begingroup$
    What have you tried so far? Have you done a)?
    $endgroup$
    – Mindlack
    Jan 6 at 17:51










  • $begingroup$
    Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
    $endgroup$
    – Antonio
    Jan 6 at 17:52












  • $begingroup$
    No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
    $endgroup$
    – Mindlack
    Jan 6 at 17:54
















$begingroup$
What have you tried so far? Have you done a)?
$endgroup$
– Mindlack
Jan 6 at 17:51




$begingroup$
What have you tried so far? Have you done a)?
$endgroup$
– Mindlack
Jan 6 at 17:51












$begingroup$
Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
$endgroup$
– Antonio
Jan 6 at 17:52






$begingroup$
Yes I have, this is the first time I have encountered this type of question, (a) I am familiar with. Feeling really dumb.
$endgroup$
– Antonio
Jan 6 at 17:52














$begingroup$
No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
$endgroup$
– Mindlack
Jan 6 at 17:54




$begingroup$
No need to feel dumb, you have to struggle with one problem of that kind and afterwards you’ll master all others! As the answer said, look for patterns. If you do not see it yet, try computing $u_5$, $u_6$, $u_7$, and so on, till you get it (and you will!).
$endgroup$
– Mindlack
Jan 6 at 17:54










1 Answer
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$begingroup$

For part (b) you need to look for patterns.
$$u_1=3$$
$$u_2=frac{2}{3}$$
$$u_3=-4$$
$$u_4=3$$
$$u_5=frac{2}{3}$$
$$u_6=-4$$
(and so on)



Do you remember how to find $i^{87}$, where $i=sqrt{-1}$? Same deal.



For part (c) you will need the identity $sum_{k=1}^nc=cn$.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    $begingroup$

    For part (b) you need to look for patterns.
    $$u_1=3$$
    $$u_2=frac{2}{3}$$
    $$u_3=-4$$
    $$u_4=3$$
    $$u_5=frac{2}{3}$$
    $$u_6=-4$$
    (and so on)



    Do you remember how to find $i^{87}$, where $i=sqrt{-1}$? Same deal.



    For part (c) you will need the identity $sum_{k=1}^nc=cn$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      For part (b) you need to look for patterns.
      $$u_1=3$$
      $$u_2=frac{2}{3}$$
      $$u_3=-4$$
      $$u_4=3$$
      $$u_5=frac{2}{3}$$
      $$u_6=-4$$
      (and so on)



      Do you remember how to find $i^{87}$, where $i=sqrt{-1}$? Same deal.



      For part (c) you will need the identity $sum_{k=1}^nc=cn$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        For part (b) you need to look for patterns.
        $$u_1=3$$
        $$u_2=frac{2}{3}$$
        $$u_3=-4$$
        $$u_4=3$$
        $$u_5=frac{2}{3}$$
        $$u_6=-4$$
        (and so on)



        Do you remember how to find $i^{87}$, where $i=sqrt{-1}$? Same deal.



        For part (c) you will need the identity $sum_{k=1}^nc=cn$.






        share|cite|improve this answer











        $endgroup$



        For part (b) you need to look for patterns.
        $$u_1=3$$
        $$u_2=frac{2}{3}$$
        $$u_3=-4$$
        $$u_4=3$$
        $$u_5=frac{2}{3}$$
        $$u_6=-4$$
        (and so on)



        Do you remember how to find $i^{87}$, where $i=sqrt{-1}$? Same deal.



        For part (c) you will need the identity $sum_{k=1}^nc=cn$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 18:02

























        answered Jan 6 at 17:50









        Ben WBen W

        2,184615




        2,184615






























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