A recurrence sequence I can't solve (A level) - Any hint is appreciated.
$begingroup$
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 ?
sequences-and-series
asked Jan 6 at 17:46
AntonioAntonio
296
$endgroup$
add a comment |
$begingroup$
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 ?
sequences-and-series
asked Jan 6 at 17:46
AntonioAntonio
296
$endgroup$
$begingroup$
What have you tried so far? Have you done a)?
$endgroup$
– Mindlack
Jan 6 at 17:51
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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
add a comment |
$begingroup$
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 ?
sequences-and-series
asked Jan 6 at 17:46
AntonioAntonio
296
$endgroup$
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
sequences-and-series
asked Jan 6 at 17:46
AntonioAntonio
296
asked Jan 6 at 17:46
AntonioAntonio
296
asked Jan 6 at 17:46
AntonioAntonio
296
asked Jan 6 at 17:46
AntonioAntonio
296
asked Jan 6 at 17:46
AntonioAntonio
296
296
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $sum_{k=1}^nc=cn$.
answered Jan 6 at 17:50
Ben WBen W
2,734918
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add a comment |
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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 $sum_{k=1}^nc=cn$.
answered Jan 6 at 17:50
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
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 $sum_{k=1}^nc=cn$.
answered Jan 6 at 17:50
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
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 $sum_{k=1}^nc=cn$.
answered Jan 6 at 17:50
Ben WBen W
2,734918
$endgroup$
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 $sum_{k=1}^nc=cn$.
answered Jan 6 at 17:50
Ben WBen W
2,734918
edited Jan 6 at 18:02
answered Jan 6 at 17:50
Ben WBen W
2,734918
answered Jan 6 at 17:50
Ben WBen W
2,734918
answered Jan 6 at 17:50
Ben WBen W
2,734918
2,734918
add a comment |
add a comment |
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$begingroup$
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