General term of $a_{n+1}=2a_n+(-1)^{n+1}$, $a_1=frac{1}{3}$ [closed]
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I tried to write a few terms and use induction, but I can't guess the general term.
sequences-and-series
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
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closed as off-topic by amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus Dec 16 '18 at 1:11
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$begingroup$
I tried to write a few terms and use induction, but I can't guess the general term.
sequences-and-series
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
$endgroup$
closed as off-topic by amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus Dec 16 '18 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I tried to write a few terms and use induction, but I can't guess the general term.
sequences-and-series
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
$endgroup$
I tried to write a few terms and use induction, but I can't guess the general term.
sequences-and-series
sequences-and-series
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
asked Dec 15 '18 at 16:18
JustAnUserJustAnUser
183
183
closed as off-topic by amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus Dec 16 '18 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus Dec 16 '18 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Nosrati, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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3 Answers
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Solution
Assume that $$a_{n+1}+p(-1)^{n+1}=2[a_n+p(-1)^n].$$
Thus $$a_{n+1}=2a_n+2p(-1)^n-p(-1)^{n+1}.$$
Comparing with $a_{n+1}=2a_n+(-1)^{n+1}$, we obtain
$$2p(-1)^n-p(-1)^{n+1}=(-1)^{n+1},$$
namely $$(3p+1)(-1)^n equiv 0,$$
which implies $p=-dfrac{1}{3}.$ Therefore
$$a_{n+1}-frac{1}{3}(-1)^{n+1}=2left[a_n-frac{1}{3}(-1)^nright].$$
We may see that $b_n:=a_n-dfrac{1}{3}(-1)^n$ is a geometrical sequence, with the initial term $b_1=dfrac{2}{3}$ and the common ratio $q=2$. Hence, for $n geq 2$,
$$a_n-frac{1}{3}(-1)^n=frac{2}{3}cdot 2^{n-1},$$
which gives us that
$$a_n=frac{2}{3}cdot 2^{n-1}+frac{1}{3}(-1)^n=frac{2^n+(-1)^n}{3},(n geq 2),$$
which also holds for $n=1$ as well.
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
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Hint: can you compute $frac{a_n}{2^n}$ using the recurrence relation?
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
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add a comment |
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Like Solving recurrence $X_n = 4X_{n-1}+5$,
Let $a_m=b_m+c(-1)^n, m=1implies b_1=a_1+c$
$$(-1)^{n+1}=a_{n+1}-2a_n=b_{n+1}+c(-1)^{n+1}-2{b_n+c(-1)^n}=b_n-2b_{n+1}+3c(-1)^{n+1}$$
Set $3c=1iff c=?$
so that $b_n=2b_{n-1}=2^rb_{n-r},0le rle n-1$
$r=n-1implies b_n=2^{n-1}b_1=?$
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solution
Assume that $$a_{n+1}+p(-1)^{n+1}=2[a_n+p(-1)^n].$$
Thus $$a_{n+1}=2a_n+2p(-1)^n-p(-1)^{n+1}.$$
Comparing with $a_{n+1}=2a_n+(-1)^{n+1}$, we obtain
$$2p(-1)^n-p(-1)^{n+1}=(-1)^{n+1},$$
namely $$(3p+1)(-1)^n equiv 0,$$
which implies $p=-dfrac{1}{3}.$ Therefore
$$a_{n+1}-frac{1}{3}(-1)^{n+1}=2left[a_n-frac{1}{3}(-1)^nright].$$
We may see that $b_n:=a_n-dfrac{1}{3}(-1)^n$ is a geometrical sequence, with the initial term $b_1=dfrac{2}{3}$ and the common ratio $q=2$. Hence, for $n geq 2$,
$$a_n-frac{1}{3}(-1)^n=frac{2}{3}cdot 2^{n-1},$$
which gives us that
$$a_n=frac{2}{3}cdot 2^{n-1}+frac{1}{3}(-1)^n=frac{2^n+(-1)^n}{3},(n geq 2),$$
which also holds for $n=1$ as well.
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
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add a comment |
$begingroup$
Solution
Assume that $$a_{n+1}+p(-1)^{n+1}=2[a_n+p(-1)^n].$$
Thus $$a_{n+1}=2a_n+2p(-1)^n-p(-1)^{n+1}.$$
Comparing with $a_{n+1}=2a_n+(-1)^{n+1}$, we obtain
$$2p(-1)^n-p(-1)^{n+1}=(-1)^{n+1},$$
namely $$(3p+1)(-1)^n equiv 0,$$
which implies $p=-dfrac{1}{3}.$ Therefore
$$a_{n+1}-frac{1}{3}(-1)^{n+1}=2left[a_n-frac{1}{3}(-1)^nright].$$
We may see that $b_n:=a_n-dfrac{1}{3}(-1)^n$ is a geometrical sequence, with the initial term $b_1=dfrac{2}{3}$ and the common ratio $q=2$. Hence, for $n geq 2$,
$$a_n-frac{1}{3}(-1)^n=frac{2}{3}cdot 2^{n-1},$$
which gives us that
$$a_n=frac{2}{3}cdot 2^{n-1}+frac{1}{3}(-1)^n=frac{2^n+(-1)^n}{3},(n geq 2),$$
which also holds for $n=1$ as well.
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
$endgroup$
add a comment |
$begingroup$
Solution
Assume that $$a_{n+1}+p(-1)^{n+1}=2[a_n+p(-1)^n].$$
Thus $$a_{n+1}=2a_n+2p(-1)^n-p(-1)^{n+1}.$$
Comparing with $a_{n+1}=2a_n+(-1)^{n+1}$, we obtain
$$2p(-1)^n-p(-1)^{n+1}=(-1)^{n+1},$$
namely $$(3p+1)(-1)^n equiv 0,$$
which implies $p=-dfrac{1}{3}.$ Therefore
$$a_{n+1}-frac{1}{3}(-1)^{n+1}=2left[a_n-frac{1}{3}(-1)^nright].$$
We may see that $b_n:=a_n-dfrac{1}{3}(-1)^n$ is a geometrical sequence, with the initial term $b_1=dfrac{2}{3}$ and the common ratio $q=2$. Hence, for $n geq 2$,
$$a_n-frac{1}{3}(-1)^n=frac{2}{3}cdot 2^{n-1},$$
which gives us that
$$a_n=frac{2}{3}cdot 2^{n-1}+frac{1}{3}(-1)^n=frac{2^n+(-1)^n}{3},(n geq 2),$$
which also holds for $n=1$ as well.
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
$endgroup$
Solution
Assume that $$a_{n+1}+p(-1)^{n+1}=2[a_n+p(-1)^n].$$
Thus $$a_{n+1}=2a_n+2p(-1)^n-p(-1)^{n+1}.$$
Comparing with $a_{n+1}=2a_n+(-1)^{n+1}$, we obtain
$$2p(-1)^n-p(-1)^{n+1}=(-1)^{n+1},$$
namely $$(3p+1)(-1)^n equiv 0,$$
which implies $p=-dfrac{1}{3}.$ Therefore
$$a_{n+1}-frac{1}{3}(-1)^{n+1}=2left[a_n-frac{1}{3}(-1)^nright].$$
We may see that $b_n:=a_n-dfrac{1}{3}(-1)^n$ is a geometrical sequence, with the initial term $b_1=dfrac{2}{3}$ and the common ratio $q=2$. Hence, for $n geq 2$,
$$a_n-frac{1}{3}(-1)^n=frac{2}{3}cdot 2^{n-1},$$
which gives us that
$$a_n=frac{2}{3}cdot 2^{n-1}+frac{1}{3}(-1)^n=frac{2^n+(-1)^n}{3},(n geq 2),$$
which also holds for $n=1$ as well.
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
edited Dec 15 '18 at 17:01
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
answered Dec 15 '18 at 16:52
mengdie1982mengdie1982
4,897618
4,897618
add a comment |
add a comment |
$begingroup$
Hint: can you compute $frac{a_n}{2^n}$ using the recurrence relation?
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
$endgroup$
add a comment |
$begingroup$
Hint: can you compute $frac{a_n}{2^n}$ using the recurrence relation?
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
$endgroup$
add a comment |
$begingroup$
Hint: can you compute $frac{a_n}{2^n}$ using the recurrence relation?
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
$endgroup$
Hint: can you compute $frac{a_n}{2^n}$ using the recurrence relation?
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
answered Dec 15 '18 at 16:20
MindlackMindlack
2,81717
2,81717
add a comment |
add a comment |
$begingroup$
Like Solving recurrence $X_n = 4X_{n-1}+5$,
Let $a_m=b_m+c(-1)^n, m=1implies b_1=a_1+c$
$$(-1)^{n+1}=a_{n+1}-2a_n=b_{n+1}+c(-1)^{n+1}-2{b_n+c(-1)^n}=b_n-2b_{n+1}+3c(-1)^{n+1}$$
Set $3c=1iff c=?$
so that $b_n=2b_{n-1}=2^rb_{n-r},0le rle n-1$
$r=n-1implies b_n=2^{n-1}b_1=?$
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Like Solving recurrence $X_n = 4X_{n-1}+5$,
Let $a_m=b_m+c(-1)^n, m=1implies b_1=a_1+c$
$$(-1)^{n+1}=a_{n+1}-2a_n=b_{n+1}+c(-1)^{n+1}-2{b_n+c(-1)^n}=b_n-2b_{n+1}+3c(-1)^{n+1}$$
Set $3c=1iff c=?$
so that $b_n=2b_{n-1}=2^rb_{n-r},0le rle n-1$
$r=n-1implies b_n=2^{n-1}b_1=?$
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Like Solving recurrence $X_n = 4X_{n-1}+5$,
Let $a_m=b_m+c(-1)^n, m=1implies b_1=a_1+c$
$$(-1)^{n+1}=a_{n+1}-2a_n=b_{n+1}+c(-1)^{n+1}-2{b_n+c(-1)^n}=b_n-2b_{n+1}+3c(-1)^{n+1}$$
Set $3c=1iff c=?$
so that $b_n=2b_{n-1}=2^rb_{n-r},0le rle n-1$
$r=n-1implies b_n=2^{n-1}b_1=?$
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Like Solving recurrence $X_n = 4X_{n-1}+5$,
Let $a_m=b_m+c(-1)^n, m=1implies b_1=a_1+c$
$$(-1)^{n+1}=a_{n+1}-2a_n=b_{n+1}+c(-1)^{n+1}-2{b_n+c(-1)^n}=b_n-2b_{n+1}+3c(-1)^{n+1}$$
Set $3c=1iff c=?$
so that $b_n=2b_{n-1}=2^rb_{n-r},0le rle n-1$
$r=n-1implies b_n=2^{n-1}b_1=?$
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 15 '18 at 16:23
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
