Mixing
A proof of a non-standard increasing alternating sum of binomial coefficients identity
$begingroup$
While working on an elementary combinatorics problem, I made the observation that the following identity holds for all examples I have calculated:
$sum_{j=2}^{n} (-1)^j (j-1) { n choose j} = 1.$
I have not been able to prove this identity. (I do not have a strong combinatorics background.) I have searched the internet for a similar identity but only came up with the standard identity
$sum_{j=0}^{n} (-1)^{j+1} (j) { n choose j} = 0$.
I have been unable to modify a proof of the standard identity to prove the one I originally stated. I also do not see an obvious (to me) way to apply the binomial theorem directly to develop a proof. Any ideas would be appreciated.
combinatorics
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
$endgroup$
add a comment |
$begingroup$
While working on an elementary combinatorics problem, I made the observation that the following identity holds for all examples I have calculated:
$sum_{j=2}^{n} (-1)^j (j-1) { n choose j} = 1.$
I have not been able to prove this identity. (I do not have a strong combinatorics background.) I have searched the internet for a similar identity but only came up with the standard identity
$sum_{j=0}^{n} (-1)^{j+1} (j) { n choose j} = 0$.
I have been unable to modify a proof of the standard identity to prove the one I originally stated. I also do not see an obvious (to me) way to apply the binomial theorem directly to develop a proof. Any ideas would be appreciated.
combinatorics
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
$endgroup$
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05
add a comment |
$begingroup$
While working on an elementary combinatorics problem, I made the observation that the following identity holds for all examples I have calculated:
$sum_{j=2}^{n} (-1)^j (j-1) { n choose j} = 1.$
I have not been able to prove this identity. (I do not have a strong combinatorics background.) I have searched the internet for a similar identity but only came up with the standard identity
$sum_{j=0}^{n} (-1)^{j+1} (j) { n choose j} = 0$.
I have been unable to modify a proof of the standard identity to prove the one I originally stated. I also do not see an obvious (to me) way to apply the binomial theorem directly to develop a proof. Any ideas would be appreciated.
combinatorics
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
$endgroup$
While working on an elementary combinatorics problem, I made the observation that the following identity holds for all examples I have calculated:
$sum_{j=2}^{n} (-1)^j (j-1) { n choose j} = 1.$
I have not been able to prove this identity. (I do not have a strong combinatorics background.) I have searched the internet for a similar identity but only came up with the standard identity
$sum_{j=0}^{n} (-1)^{j+1} (j) { n choose j} = 0$.
I have been unable to modify a proof of the standard identity to prove the one I originally stated. I also do not see an obvious (to me) way to apply the binomial theorem directly to develop a proof. Any ideas would be appreciated.
combinatorics
combinatorics
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
edited Feb 3 at 3:05
JeanDeBourque
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
asked Feb 2 at 20:13
JeanDeBourqueJeanDeBourque
133
133
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05
add a comment |
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
begin{align}
sum_{j=2}^n (-1)^j (j-1){nchoose j}&= 1 + sum_{j=0}^n (-1)^j (j-1){nchoose j}\
&= 1 + sum_{j=0}^n (-1)^j j{nchoose j} - sum_{j=0}^n (-1)^j {nchoose j} \
&= 1 + 0 + (1 + (-1))^n = 1.
end{align}
$$
using the "standard identity" that you stated and the binomial theorem.
answered Feb 3 at 1:18
irchansirchans
1,22949
$endgroup$
add a comment |
$begingroup$
EDIT: As Mike Earnest noted in the comments, this works for $(-1)^j$, not $(-1)^n$, which I didn't notice the question actually said. I'm assuming that OP meant the former, since the latter is not true.
I have a semi-combinatorial proof. We interpret the expression as counting pairs $(S,x)$, where $Ssubseteq [n]$ where $[n]={1,2,ldots,n}$, such that $xin S$, and $x$ is not the largest element of $x$. That is, one element of the set, which is not the largest one, is deemed special. This is because ${nchoose j}$ chooses a size $j$ subset of $[n]$, and $(j-1)$ chooses an element, besides the largest one. Each term has a positive contribution if $|S|$ is even, and a negative contribution if $|S|$ is odd. Then we want to show that the number of such pairs $(S,x)$ where $|S|$ is even is one greater than the number of pairs $(S,x)$ where $|S|$ is odd.
We prove this by bijection- kind of. Given $Ssubseteq [n]$, consider the following function:
$f(S)=begin{cases}
Scup{n}, text{if }nnotin S\
Ssetminus{n},text{if }nin S
end{cases}$.
Then given a pair $(S,x)$, consider the pair $g(S,x)=(f(S),x)$. Then $f(S)$ is even when $S$ is odd, and vice versa. This is similar to the proof that the number of even subsets of $[n]$ is equal to the number of odd subsets of $[n]$ when $n>0$. Then $g(S,x)$ is matching the pairs $(S,x)$ where $|S|$ is even with the pairs where $|S|$ is odd. But remember that we want $xin S$ to not be the greatest element of $S$. If $nin S$ and $x$ is the second greatest element of $S$, then $(f(S),x)$ will not satisfy the given requirement.
We've reduced the problem into looking at this special case where $nin S$ and $x$ is the second greatest element of $S$. How many such pairs are there? Counting such pairs corresponds to choosing a subset $T$ of $[n-1]$ and automatically picking its greatest element $y$, since we can then just take $(Tcup{n},y)$ to be our pair. However, we can't take $T=varnothing$, since it does not have a greatest element. In other words, we just look at $#text{odd subsets of } [n-1]-#text{nonempty even subsets of }[n-1]$ (since if $T$ is odd, $Tcup{n}$ is even, and vice versa), which we know is $1$, since $#text{odd subsets of }[n-1]-#text{even subsets of }[n-1]=0$, but not subtracting $1$ for the empty set increases the value by $1$, which is where we get the $1$ from.
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097744%2fa-proof-of-a-non-standard-increasing-alternating-sum-of-binomial-coefficients-id%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
begin{align}
sum_{j=2}^n (-1)^j (j-1){nchoose j}&= 1 + sum_{j=0}^n (-1)^j (j-1){nchoose j}\
&= 1 + sum_{j=0}^n (-1)^j j{nchoose j} - sum_{j=0}^n (-1)^j {nchoose j} \
&= 1 + 0 + (1 + (-1))^n = 1.
end{align}
$$
using the "standard identity" that you stated and the binomial theorem.
answered Feb 3 at 1:18
irchansirchans
1,22949
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
sum_{j=2}^n (-1)^j (j-1){nchoose j}&= 1 + sum_{j=0}^n (-1)^j (j-1){nchoose j}\
&= 1 + sum_{j=0}^n (-1)^j j{nchoose j} - sum_{j=0}^n (-1)^j {nchoose j} \
&= 1 + 0 + (1 + (-1))^n = 1.
end{align}
$$
using the "standard identity" that you stated and the binomial theorem.
answered Feb 3 at 1:18
irchansirchans
1,22949
$endgroup$
add a comment |
$begingroup$
$$
begin{align}
sum_{j=2}^n (-1)^j (j-1){nchoose j}&= 1 + sum_{j=0}^n (-1)^j (j-1){nchoose j}\
&= 1 + sum_{j=0}^n (-1)^j j{nchoose j} - sum_{j=0}^n (-1)^j {nchoose j} \
&= 1 + 0 + (1 + (-1))^n = 1.
end{align}
$$
using the "standard identity" that you stated and the binomial theorem.
answered Feb 3 at 1:18
irchansirchans
1,22949
$endgroup$
$$
begin{align}
sum_{j=2}^n (-1)^j (j-1){nchoose j}&= 1 + sum_{j=0}^n (-1)^j (j-1){nchoose j}\
&= 1 + sum_{j=0}^n (-1)^j j{nchoose j} - sum_{j=0}^n (-1)^j {nchoose j} \
&= 1 + 0 + (1 + (-1))^n = 1.
end{align}
$$
using the "standard identity" that you stated and the binomial theorem.
answered Feb 3 at 1:18
irchansirchans
1,22949
answered Feb 3 at 1:18
irchansirchans
1,22949
answered Feb 3 at 1:18
irchansirchans
1,22949
answered Feb 3 at 1:18
irchansirchans
1,22949
1,22949
add a comment |
add a comment |
$begingroup$
EDIT: As Mike Earnest noted in the comments, this works for $(-1)^j$, not $(-1)^n$, which I didn't notice the question actually said. I'm assuming that OP meant the former, since the latter is not true.
I have a semi-combinatorial proof. We interpret the expression as counting pairs $(S,x)$, where $Ssubseteq [n]$ where $[n]={1,2,ldots,n}$, such that $xin S$, and $x$ is not the largest element of $x$. That is, one element of the set, which is not the largest one, is deemed special. This is because ${nchoose j}$ chooses a size $j$ subset of $[n]$, and $(j-1)$ chooses an element, besides the largest one. Each term has a positive contribution if $|S|$ is even, and a negative contribution if $|S|$ is odd. Then we want to show that the number of such pairs $(S,x)$ where $|S|$ is even is one greater than the number of pairs $(S,x)$ where $|S|$ is odd.
We prove this by bijection- kind of. Given $Ssubseteq [n]$, consider the following function:
$f(S)=begin{cases}
Scup{n}, text{if }nnotin S\
Ssetminus{n},text{if }nin S
end{cases}$.
Then given a pair $(S,x)$, consider the pair $g(S,x)=(f(S),x)$. Then $f(S)$ is even when $S$ is odd, and vice versa. This is similar to the proof that the number of even subsets of $[n]$ is equal to the number of odd subsets of $[n]$ when $n>0$. Then $g(S,x)$ is matching the pairs $(S,x)$ where $|S|$ is even with the pairs where $|S|$ is odd. But remember that we want $xin S$ to not be the greatest element of $S$. If $nin S$ and $x$ is the second greatest element of $S$, then $(f(S),x)$ will not satisfy the given requirement.
We've reduced the problem into looking at this special case where $nin S$ and $x$ is the second greatest element of $S$. How many such pairs are there? Counting such pairs corresponds to choosing a subset $T$ of $[n-1]$ and automatically picking its greatest element $y$, since we can then just take $(Tcup{n},y)$ to be our pair. However, we can't take $T=varnothing$, since it does not have a greatest element. In other words, we just look at $#text{odd subsets of } [n-1]-#text{nonempty even subsets of }[n-1]$ (since if $T$ is odd, $Tcup{n}$ is even, and vice versa), which we know is $1$, since $#text{odd subsets of }[n-1]-#text{even subsets of }[n-1]=0$, but not subtracting $1$ for the empty set increases the value by $1$, which is where we get the $1$ from.
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
$endgroup$
add a comment |
$begingroup$
EDIT: As Mike Earnest noted in the comments, this works for $(-1)^j$, not $(-1)^n$, which I didn't notice the question actually said. I'm assuming that OP meant the former, since the latter is not true.
I have a semi-combinatorial proof. We interpret the expression as counting pairs $(S,x)$, where $Ssubseteq [n]$ where $[n]={1,2,ldots,n}$, such that $xin S$, and $x$ is not the largest element of $x$. That is, one element of the set, which is not the largest one, is deemed special. This is because ${nchoose j}$ chooses a size $j$ subset of $[n]$, and $(j-1)$ chooses an element, besides the largest one. Each term has a positive contribution if $|S|$ is even, and a negative contribution if $|S|$ is odd. Then we want to show that the number of such pairs $(S,x)$ where $|S|$ is even is one greater than the number of pairs $(S,x)$ where $|S|$ is odd.
We prove this by bijection- kind of. Given $Ssubseteq [n]$, consider the following function:
$f(S)=begin{cases}
Scup{n}, text{if }nnotin S\
Ssetminus{n},text{if }nin S
end{cases}$.
Then given a pair $(S,x)$, consider the pair $g(S,x)=(f(S),x)$. Then $f(S)$ is even when $S$ is odd, and vice versa. This is similar to the proof that the number of even subsets of $[n]$ is equal to the number of odd subsets of $[n]$ when $n>0$. Then $g(S,x)$ is matching the pairs $(S,x)$ where $|S|$ is even with the pairs where $|S|$ is odd. But remember that we want $xin S$ to not be the greatest element of $S$. If $nin S$ and $x$ is the second greatest element of $S$, then $(f(S),x)$ will not satisfy the given requirement.
We've reduced the problem into looking at this special case where $nin S$ and $x$ is the second greatest element of $S$. How many such pairs are there? Counting such pairs corresponds to choosing a subset $T$ of $[n-1]$ and automatically picking its greatest element $y$, since we can then just take $(Tcup{n},y)$ to be our pair. However, we can't take $T=varnothing$, since it does not have a greatest element. In other words, we just look at $#text{odd subsets of } [n-1]-#text{nonempty even subsets of }[n-1]$ (since if $T$ is odd, $Tcup{n}$ is even, and vice versa), which we know is $1$, since $#text{odd subsets of }[n-1]-#text{even subsets of }[n-1]=0$, but not subtracting $1$ for the empty set increases the value by $1$, which is where we get the $1$ from.
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
$endgroup$
add a comment |
$begingroup$
EDIT: As Mike Earnest noted in the comments, this works for $(-1)^j$, not $(-1)^n$, which I didn't notice the question actually said. I'm assuming that OP meant the former, since the latter is not true.
I have a semi-combinatorial proof. We interpret the expression as counting pairs $(S,x)$, where $Ssubseteq [n]$ where $[n]={1,2,ldots,n}$, such that $xin S$, and $x$ is not the largest element of $x$. That is, one element of the set, which is not the largest one, is deemed special. This is because ${nchoose j}$ chooses a size $j$ subset of $[n]$, and $(j-1)$ chooses an element, besides the largest one. Each term has a positive contribution if $|S|$ is even, and a negative contribution if $|S|$ is odd. Then we want to show that the number of such pairs $(S,x)$ where $|S|$ is even is one greater than the number of pairs $(S,x)$ where $|S|$ is odd.
We prove this by bijection- kind of. Given $Ssubseteq [n]$, consider the following function:
$f(S)=begin{cases}
Scup{n}, text{if }nnotin S\
Ssetminus{n},text{if }nin S
end{cases}$.
Then given a pair $(S,x)$, consider the pair $g(S,x)=(f(S),x)$. Then $f(S)$ is even when $S$ is odd, and vice versa. This is similar to the proof that the number of even subsets of $[n]$ is equal to the number of odd subsets of $[n]$ when $n>0$. Then $g(S,x)$ is matching the pairs $(S,x)$ where $|S|$ is even with the pairs where $|S|$ is odd. But remember that we want $xin S$ to not be the greatest element of $S$. If $nin S$ and $x$ is the second greatest element of $S$, then $(f(S),x)$ will not satisfy the given requirement.
We've reduced the problem into looking at this special case where $nin S$ and $x$ is the second greatest element of $S$. How many such pairs are there? Counting such pairs corresponds to choosing a subset $T$ of $[n-1]$ and automatically picking its greatest element $y$, since we can then just take $(Tcup{n},y)$ to be our pair. However, we can't take $T=varnothing$, since it does not have a greatest element. In other words, we just look at $#text{odd subsets of } [n-1]-#text{nonempty even subsets of }[n-1]$ (since if $T$ is odd, $Tcup{n}$ is even, and vice versa), which we know is $1$, since $#text{odd subsets of }[n-1]-#text{even subsets of }[n-1]=0$, but not subtracting $1$ for the empty set increases the value by $1$, which is where we get the $1$ from.
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
$endgroup$
EDIT: As Mike Earnest noted in the comments, this works for $(-1)^j$, not $(-1)^n$, which I didn't notice the question actually said. I'm assuming that OP meant the former, since the latter is not true.
I have a semi-combinatorial proof. We interpret the expression as counting pairs $(S,x)$, where $Ssubseteq [n]$ where $[n]={1,2,ldots,n}$, such that $xin S$, and $x$ is not the largest element of $x$. That is, one element of the set, which is not the largest one, is deemed special. This is because ${nchoose j}$ chooses a size $j$ subset of $[n]$, and $(j-1)$ chooses an element, besides the largest one. Each term has a positive contribution if $|S|$ is even, and a negative contribution if $|S|$ is odd. Then we want to show that the number of such pairs $(S,x)$ where $|S|$ is even is one greater than the number of pairs $(S,x)$ where $|S|$ is odd.
We prove this by bijection- kind of. Given $Ssubseteq [n]$, consider the following function:
$f(S)=begin{cases}
Scup{n}, text{if }nnotin S\
Ssetminus{n},text{if }nin S
end{cases}$.
Then given a pair $(S,x)$, consider the pair $g(S,x)=(f(S),x)$. Then $f(S)$ is even when $S$ is odd, and vice versa. This is similar to the proof that the number of even subsets of $[n]$ is equal to the number of odd subsets of $[n]$ when $n>0$. Then $g(S,x)$ is matching the pairs $(S,x)$ where $|S|$ is even with the pairs where $|S|$ is odd. But remember that we want $xin S$ to not be the greatest element of $S$. If $nin S$ and $x$ is the second greatest element of $S$, then $(f(S),x)$ will not satisfy the given requirement.
We've reduced the problem into looking at this special case where $nin S$ and $x$ is the second greatest element of $S$. How many such pairs are there? Counting such pairs corresponds to choosing a subset $T$ of $[n-1]$ and automatically picking its greatest element $y$, since we can then just take $(Tcup{n},y)$ to be our pair. However, we can't take $T=varnothing$, since it does not have a greatest element. In other words, we just look at $#text{odd subsets of } [n-1]-#text{nonempty even subsets of }[n-1]$ (since if $T$ is odd, $Tcup{n}$ is even, and vice versa), which we know is $1$, since $#text{odd subsets of }[n-1]-#text{even subsets of }[n-1]=0$, but not subtracting $1$ for the empty set increases the value by $1$, which is where we get the $1$ from.
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
edited Feb 2 at 21:28
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
answered Feb 2 at 21:00
Kevin LongKevin Long
3,58121431
3,58121431
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097744%2fa-proof-of-a-non-standard-increasing-alternating-sum-of-binomial-coefficients-id%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Also, do you mean $(-1)^j$ inside your summation? With $(-1)^n$, the identity is false.
$endgroup$
– Mike Earnest
Feb 2 at 21:03
$begingroup$
Yes I did type a $n$ for the exponent of (-1) when it should have been $j$. My apologies.
$endgroup$
– JeanDeBourque
Feb 3 at 3:06
$begingroup$
Thank you for the responses.
$endgroup$
– JeanDeBourque
Feb 3 at 23:05