Mixing
Sum of binomial coefficients, with alternate signs, congruent two or zero modulo four
$begingroup$
I've come across the following formula:
$$
S(n) ;:=
sum
_{
begin{array}{c}
k=0 \
k equiv_4 0,; 2
end{array}
}
^{
n
}
2 cdot
binom{n}{k}cdot
(-1)^{k over 2} $$
where the summation run over the values of $k$ in $[0, n]$ that are congruent modulo 4 to 0 or 2.
Can this be simplified?
I've numerically verified that the sum takes the following values:
$$
S(n) =
begin{cases}
(-1)^{n over 4} cdot 2^{{nover 2} + 1}, & text{if $n equiv_4 0 $} \
S(n-1) , & text{if $n equiv_4 1 $} \
0, & text{if $n equiv_4 2 $} \
(-1)^{n +1 over 4} cdot 2^{n+1over2}, & text{if $n equiv_4 3 $} \
end{cases}
$$
but except for the case $n equiv_4 2 $ I can see why this is the case.
Particularly interesting, to me, are the fact that this sum is either 0 or a power of two and the case $n equiv_4 1 $ where the two partial sums of the binomial coefficients in adjacent rows total the same (e.g. $ 2 cdot 1 - 2 cdot 6 + 2 cdot 1 = 2 cdot 1 - 2 cdot 10 + 2 cdot 5$ for the 5th and 6th row of the Pascal's triangle).
summation proof-writing binomial-coefficients
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
$endgroup$
add a comment |
$begingroup$
I've come across the following formula:
$$
S(n) ;:=
sum
_{
begin{array}{c}
k=0 \
k equiv_4 0,; 2
end{array}
}
^{
n
}
2 cdot
binom{n}{k}cdot
(-1)^{k over 2} $$
where the summation run over the values of $k$ in $[0, n]$ that are congruent modulo 4 to 0 or 2.
Can this be simplified?
I've numerically verified that the sum takes the following values:
$$
S(n) =
begin{cases}
(-1)^{n over 4} cdot 2^{{nover 2} + 1}, & text{if $n equiv_4 0 $} \
S(n-1) , & text{if $n equiv_4 1 $} \
0, & text{if $n equiv_4 2 $} \
(-1)^{n +1 over 4} cdot 2^{n+1over2}, & text{if $n equiv_4 3 $} \
end{cases}
$$
but except for the case $n equiv_4 2 $ I can see why this is the case.
Particularly interesting, to me, are the fact that this sum is either 0 or a power of two and the case $n equiv_4 1 $ where the two partial sums of the binomial coefficients in adjacent rows total the same (e.g. $ 2 cdot 1 - 2 cdot 6 + 2 cdot 1 = 2 cdot 1 - 2 cdot 10 + 2 cdot 5$ for the 5th and 6th row of the Pascal's triangle).
summation proof-writing binomial-coefficients
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
$endgroup$
2
$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42
add a comment |
$begingroup$
I've come across the following formula:
$$
S(n) ;:=
sum
_{
begin{array}{c}
k=0 \
k equiv_4 0,; 2
end{array}
}
^{
n
}
2 cdot
binom{n}{k}cdot
(-1)^{k over 2} $$
where the summation run over the values of $k$ in $[0, n]$ that are congruent modulo 4 to 0 or 2.
Can this be simplified?
I've numerically verified that the sum takes the following values:
$$
S(n) =
begin{cases}
(-1)^{n over 4} cdot 2^{{nover 2} + 1}, & text{if $n equiv_4 0 $} \
S(n-1) , & text{if $n equiv_4 1 $} \
0, & text{if $n equiv_4 2 $} \
(-1)^{n +1 over 4} cdot 2^{n+1over2}, & text{if $n equiv_4 3 $} \
end{cases}
$$
but except for the case $n equiv_4 2 $ I can see why this is the case.
Particularly interesting, to me, are the fact that this sum is either 0 or a power of two and the case $n equiv_4 1 $ where the two partial sums of the binomial coefficients in adjacent rows total the same (e.g. $ 2 cdot 1 - 2 cdot 6 + 2 cdot 1 = 2 cdot 1 - 2 cdot 10 + 2 cdot 5$ for the 5th and 6th row of the Pascal's triangle).
summation proof-writing binomial-coefficients
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
$endgroup$
I've come across the following formula:
$$
S(n) ;:=
sum
_{
begin{array}{c}
k=0 \
k equiv_4 0,; 2
end{array}
}
^{
n
}
2 cdot
binom{n}{k}cdot
(-1)^{k over 2} $$
where the summation run over the values of $k$ in $[0, n]$ that are congruent modulo 4 to 0 or 2.
Can this be simplified?
I've numerically verified that the sum takes the following values:
$$
S(n) =
begin{cases}
(-1)^{n over 4} cdot 2^{{nover 2} + 1}, & text{if $n equiv_4 0 $} \
S(n-1) , & text{if $n equiv_4 1 $} \
0, & text{if $n equiv_4 2 $} \
(-1)^{n +1 over 4} cdot 2^{n+1over2}, & text{if $n equiv_4 3 $} \
end{cases}
$$
but except for the case $n equiv_4 2 $ I can see why this is the case.
Particularly interesting, to me, are the fact that this sum is either 0 or a power of two and the case $n equiv_4 1 $ where the two partial sums of the binomial coefficients in adjacent rows total the same (e.g. $ 2 cdot 1 - 2 cdot 6 + 2 cdot 1 = 2 cdot 1 - 2 cdot 10 + 2 cdot 5$ for the 5th and 6th row of the Pascal's triangle).
summation proof-writing binomial-coefficients
summation proof-writing binomial-coefficients
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
asked Jan 9 at 17:34
Margaret BloomMargaret Bloom
1183
1183
2
$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42
add a comment |
2
$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42
2
2
$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42
$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$S(n)=2sum_{kequiv_40,2}^n(-1)^{k/2}binom nk=2Big(binom n0-binom n2+binom n4...Big)$$
Consider $(1+x)^n=binom n0+binom n1x+binom n2x^2+...+binom nnx^n$.
Substitute $x=i$ to get$$(1+i)^n=binom n0+binom n1i-binom n2-binom n3i+binom n4...$$
This gives$$mathfrak R((1+i)^n)=frac{S(n)}2\therefore S(n)=2mathfrak R(2^{n/2}e^{inpi/4})=2^{n/2+1}cos(npi/4)$$
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
$$S(n)=2sum_{kequiv_40,2}^n(-1)^{k/2}binom nk=2Big(binom n0-binom n2+binom n4...Big)$$
Consider $(1+x)^n=binom n0+binom n1x+binom n2x^2+...+binom nnx^n$.
Substitute $x=i$ to get$$(1+i)^n=binom n0+binom n1i-binom n2-binom n3i+binom n4...$$
This gives$$mathfrak R((1+i)^n)=frac{S(n)}2\therefore S(n)=2mathfrak R(2^{n/2}e^{inpi/4})=2^{n/2+1}cos(npi/4)$$
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
$$S(n)=2sum_{kequiv_40,2}^n(-1)^{k/2}binom nk=2Big(binom n0-binom n2+binom n4...Big)$$
Consider $(1+x)^n=binom n0+binom n1x+binom n2x^2+...+binom nnx^n$.
Substitute $x=i$ to get$$(1+i)^n=binom n0+binom n1i-binom n2-binom n3i+binom n4...$$
This gives$$mathfrak R((1+i)^n)=frac{S(n)}2\therefore S(n)=2mathfrak R(2^{n/2}e^{inpi/4})=2^{n/2+1}cos(npi/4)$$
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
$$S(n)=2sum_{kequiv_40,2}^n(-1)^{k/2}binom nk=2Big(binom n0-binom n2+binom n4...Big)$$
Consider $(1+x)^n=binom n0+binom n1x+binom n2x^2+...+binom nnx^n$.
Substitute $x=i$ to get$$(1+i)^n=binom n0+binom n1i-binom n2-binom n3i+binom n4...$$
This gives$$mathfrak R((1+i)^n)=frac{S(n)}2\therefore S(n)=2mathfrak R(2^{n/2}e^{inpi/4})=2^{n/2+1}cos(npi/4)$$
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
$endgroup$
$$S(n)=2sum_{kequiv_40,2}^n(-1)^{k/2}binom nk=2Big(binom n0-binom n2+binom n4...Big)$$
Consider $(1+x)^n=binom n0+binom n1x+binom n2x^2+...+binom nnx^n$.
Substitute $x=i$ to get$$(1+i)^n=binom n0+binom n1i-binom n2-binom n3i+binom n4...$$
This gives$$mathfrak R((1+i)^n)=frac{S(n)}2\therefore S(n)=2mathfrak R(2^{n/2}e^{inpi/4})=2^{n/2+1}cos(npi/4)$$
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
answered Jan 9 at 18:49
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
add a comment |
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$begingroup$
Two possible simplifications. The sum is over all even $k$; no need to think mod $4$ about the index. You can factor $2$ from the sum.
$endgroup$
– Ethan Bolker
Jan 9 at 17:42