The nth term of the maclaurin sequence of $frac 1{1+x+x^2}$












4












$begingroup$


$$frac 1{1+x+x^2}$$
$$ = sum^infty_{n=0} {(-1)}^n{(x+x^2)}^n$$
$$ = sum^infty_{n=0}{(-x)}^n sum^n_{k=0} {_nC_k}x^k$$
$$ = sum^infty_{n=0}sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$$



I wanted to simplify the last expression to an expression with a single $sum$.



So I tried substituting a few terms.



$$begin{align}
tag{n=6} sum^3_{k=0}{(-1)}^{6-k}{_{6-k}C_k} = 1 \
tag{n=7} sum^3_{k=0}{(-1)}^{7-k}{_{7-k}C_k} = -1 \
tag{n=8} sum^4_{k=0}{(-1)}^{8-k}{_{8-k}C_k} = 0 \
tag{n=9} sum^4_{k=0}{(-1)}^{9-k}{_{9-k}C_k} = 1 \
end{align} $$



Then, is there any generic simplified form of $sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$?






sequences-and-series combinatorics binomial-coefficients euler-maclaurin







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  • 1




    $begingroup$
    See also math.stackexchange.com/questions/3053514/… for a proof.
    $endgroup$
    – Mike Earnest
    Jan 12 at 18:33
















4












$begingroup$


$$frac 1{1+x+x^2}$$
$$ = sum^infty_{n=0} {(-1)}^n{(x+x^2)}^n$$
$$ = sum^infty_{n=0}{(-x)}^n sum^n_{k=0} {_nC_k}x^k$$
$$ = sum^infty_{n=0}sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$$



I wanted to simplify the last expression to an expression with a single $sum$.



So I tried substituting a few terms.



$$begin{align}
tag{n=6} sum^3_{k=0}{(-1)}^{6-k}{_{6-k}C_k} = 1 \
tag{n=7} sum^3_{k=0}{(-1)}^{7-k}{_{7-k}C_k} = -1 \
tag{n=8} sum^4_{k=0}{(-1)}^{8-k}{_{8-k}C_k} = 0 \
tag{n=9} sum^4_{k=0}{(-1)}^{9-k}{_{9-k}C_k} = 1 \
end{align} $$



Then, is there any generic simplified form of $sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$?






sequences-and-series combinatorics binomial-coefficients euler-maclaurin







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See also math.stackexchange.com/questions/3053514/… for a proof.
    $endgroup$
    – Mike Earnest
    Jan 12 at 18:33














4












4








4


0



$begingroup$


$$frac 1{1+x+x^2}$$
$$ = sum^infty_{n=0} {(-1)}^n{(x+x^2)}^n$$
$$ = sum^infty_{n=0}{(-x)}^n sum^n_{k=0} {_nC_k}x^k$$
$$ = sum^infty_{n=0}sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$$



I wanted to simplify the last expression to an expression with a single $sum$.



So I tried substituting a few terms.



$$begin{align}
tag{n=6} sum^3_{k=0}{(-1)}^{6-k}{_{6-k}C_k} = 1 \
tag{n=7} sum^3_{k=0}{(-1)}^{7-k}{_{7-k}C_k} = -1 \
tag{n=8} sum^4_{k=0}{(-1)}^{8-k}{_{8-k}C_k} = 0 \
tag{n=9} sum^4_{k=0}{(-1)}^{9-k}{_{9-k}C_k} = 1 \
end{align} $$



Then, is there any generic simplified form of $sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$?






sequences-and-series combinatorics binomial-coefficients euler-maclaurin







share|cite|improve this question









$endgroup$




$$frac 1{1+x+x^2}$$
$$ = sum^infty_{n=0} {(-1)}^n{(x+x^2)}^n$$
$$ = sum^infty_{n=0}{(-x)}^n sum^n_{k=0} {_nC_k}x^k$$
$$ = sum^infty_{n=0}sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$$



I wanted to simplify the last expression to an expression with a single $sum$.



So I tried substituting a few terms.



$$begin{align}
tag{n=6} sum^3_{k=0}{(-1)}^{6-k}{_{6-k}C_k} = 1 \
tag{n=7} sum^3_{k=0}{(-1)}^{7-k}{_{7-k}C_k} = -1 \
tag{n=8} sum^4_{k=0}{(-1)}^{8-k}{_{8-k}C_k} = 0 \
tag{n=9} sum^4_{k=0}{(-1)}^{9-k}{_{9-k}C_k} = 1 \
end{align} $$



Then, is there any generic simplified form of $sum^{[frac n2]}_{k=0}{(-1)}^{n-k}{_{n-k}C_k}x^n$?






sequences-and-series combinatorics binomial-coefficients euler-maclaurin




sequences-and-series combinatorics binomial-coefficients euler-maclaurin






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asked Jan 12 at 13:29









KYHSGeekCodeKYHSGeekCode

334112




334112








  • 1




    $begingroup$
    See also math.stackexchange.com/questions/3053514/… for a proof.
    $endgroup$
    – Mike Earnest
    Jan 12 at 18:33














  • 1




    $begingroup$
    See also math.stackexchange.com/questions/3053514/… for a proof.
    $endgroup$
    – Mike Earnest
    Jan 12 at 18:33








1




1




$begingroup$
See also math.stackexchange.com/questions/3053514/… for a proof.
$endgroup$
– Mike Earnest
Jan 12 at 18:33




$begingroup$
See also math.stackexchange.com/questions/3053514/… for a proof.
$endgroup$
– Mike Earnest
Jan 12 at 18:33










1 Answer
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$begingroup$

Hint for a simpler approach. Note that for $|x|<1$,
$$frac{1}{1+x+x^2}=frac{1-x}{1-x^3}=(1-x)sum_{k=0}^{infty} x^{3k}.$$
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Of course, You are a genius!!!! Thank you
    $endgroup$
    – KYHSGeekCode
    Jan 12 at 13:47











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1 Answer
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active

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active

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10












$begingroup$

Hint for a simpler approach. Note that for $|x|<1$,
$$frac{1}{1+x+x^2}=frac{1-x}{1-x^3}=(1-x)sum_{k=0}^{infty} x^{3k}.$$
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Of course, You are a genius!!!! Thank you
    $endgroup$
    – KYHSGeekCode
    Jan 12 at 13:47
















10












$begingroup$

Hint for a simpler approach. Note that for $|x|<1$,
$$frac{1}{1+x+x^2}=frac{1-x}{1-x^3}=(1-x)sum_{k=0}^{infty} x^{3k}.$$
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Of course, You are a genius!!!! Thank you
    $endgroup$
    – KYHSGeekCode
    Jan 12 at 13:47














10












10








10





$begingroup$

Hint for a simpler approach. Note that for $|x|<1$,
$$frac{1}{1+x+x^2}=frac{1-x}{1-x^3}=(1-x)sum_{k=0}^{infty} x^{3k}.$$
Can you take it from here?






share|cite|improve this answer











$endgroup$



Hint for a simpler approach. Note that for $|x|<1$,
$$frac{1}{1+x+x^2}=frac{1-x}{1-x^3}=(1-x)sum_{k=0}^{infty} x^{3k}.$$
Can you take it from here?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 13:43

























answered Jan 12 at 13:36









Robert ZRobert Z

97.1k1066137




97.1k1066137








  • 1




    $begingroup$
    Of course, You are a genius!!!! Thank you
    $endgroup$
    – KYHSGeekCode
    Jan 12 at 13:47














  • 1




    $begingroup$
    Of course, You are a genius!!!! Thank you
    $endgroup$
    – KYHSGeekCode
    Jan 12 at 13:47








1




1




$begingroup$
Of course, You are a genius!!!! Thank you
$endgroup$
– KYHSGeekCode
Jan 12 at 13:47




$begingroup$
Of course, You are a genius!!!! Thank you
$endgroup$
– KYHSGeekCode
Jan 12 at 13:47


















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