Could someone please show me the steps of this sum?Markov Chain: I don't understand this solution to a conditional probability problem after n state transitions…Using binomial theorem to evaluate summation $sum_k=0^n frac1k+1 binom nk$ in closed formHaving trouble find sum of this equation which has exponents. Can someone please help me solve thisCan somone help me do this double sum problem. I know how to do it manually, but I would like to know how to do it using summation formulas.Why is $sum_j = 1^n dfrac1n = 1$Converting a double summation to a single summationSolving Triple Summation Cases
What is the need of methods like GET and POST in the HTTP protocol?
The 100 soldier problem
Circle divided by lines between a blue dots
Can Northern Ireland's border issue be solved by repartition?
Why are Fuji lenses more expensive than others?
Safely hang a mirror that does not have hooks
I feel like most of my characters are the same, what can I do?
Is it true that, "just ten trading days represent 63 per cent of the returns of the past 50 years"?
What do these pins mean? Where should I plug them in?
What was an "insurance cover"?
As a discovery writer, how do I complete an unfinished novel (which has highly diverged from the original plot ) after a time-gap?
CDG baggage claim before or after immigration?
Which museums have artworks of all four ninja turtles' namesakes?
Pseudo Game of Cups in Python
Are there hydrocarbons on the Moon?
How do I improve in sight reading?
Social leper versus social leopard
How can I get a language selector in top bar in Ubuntu 19.04?
Apple Developer Program Refund Help
How would a native speaker correct themselves when they misspeak?
GitHub repo with Apache License version 2 in package.json, but no full license copy nor comment headers
Algorithm that spans orthogonal vectors: Python
Can multiple wall timers turn lights on or off when required?
Debussy as term for bathroom?
Could someone please show me the steps of this sum?
Markov Chain: I don't understand this solution to a conditional probability problem after n state transitions…Using binomial theorem to evaluate summation $sum_k=0^n frac1k+1 binom nk$ in closed formHaving trouble find sum of this equation which has exponents. Can someone please help me solve thisCan somone help me do this double sum problem. I know how to do it manually, but I would like to know how to do it using summation formulas.Why is $sum_j = 1^n dfrac1n = 1$Converting a double summation to a single summationSolving Triple Summation Cases
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
In my probability textbook, I saw this summation, $sum_n=0^N$$Nchoosen$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!
summation
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
$endgroup$
add a comment
|
$begingroup$
In my probability textbook, I saw this summation, $sum_n=0^N$$Nchoosen$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!
summation
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
$endgroup$
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago
add a comment
|
$begingroup$
In my probability textbook, I saw this summation, $sum_n=0^N$$Nchoosen$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!
summation
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
$endgroup$
In my probability textbook, I saw this summation, $sum_n=0^N$$Nchoosen$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!
summation
summation
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
edited 6 hours ago
JustWandering
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
asked 11 hours ago
JustWanderingJustWandering
50310 bronze badges
50310 bronze badges
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago
add a comment
|
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$)
$$ sum_n=0^1 binom1n s^n = 1+s=(1+s)^1$$
($N=m to N=m+1$)
beginalign* (1+s)^N+1 &= (1+s)(1+s)^N=(1+s)sum_n=0^N binomNn s^n \&= sum_n=0^N binomNn s^n+sum_n=0^N binomNn s^n+1 \ &= 1+sum_n=1^N binomNn s^n + sum_n=1^N binomNn-1 s^n + s^N+1 \ &= 1+sum_n=1^N left( binomNn + binomNn-1 right) s^n + s^N+1 \ &= 1+sum_n=1^N binomN+1n s^n + s^N+1 \&
= sum_n=0^N+1 binomN+1n s^n
endalign*
Here I use the recursive formula $$ binomnk = binomn-1k+binomn-1k-1 $$
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
$endgroup$
add a comment
|
$begingroup$
By the Binomial Theorem
$$(a+b)^N=sum_n=0^NNchoose na^ncdot b^N-n$$
choosing $a=s$ and $b=1$ leads to
$$(1+s)^N=(s+1)^N=sum_n=0^NNchoose ns^ncdot 1^N-n=sum_n=0^NNchoose ns^n$$
$endgroup$
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/4.0/"u003ecc by-sa 4.0 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%2f3360862%2fcould-someone-please-show-me-the-steps-of-this-sum%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
$begingroup$
If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$)
$$ sum_n=0^1 binom1n s^n = 1+s=(1+s)^1$$
($N=m to N=m+1$)
beginalign* (1+s)^N+1 &= (1+s)(1+s)^N=(1+s)sum_n=0^N binomNn s^n \&= sum_n=0^N binomNn s^n+sum_n=0^N binomNn s^n+1 \ &= 1+sum_n=1^N binomNn s^n + sum_n=1^N binomNn-1 s^n + s^N+1 \ &= 1+sum_n=1^N left( binomNn + binomNn-1 right) s^n + s^N+1 \ &= 1+sum_n=1^N binomN+1n s^n + s^N+1 \&
= sum_n=0^N+1 binomN+1n s^n
endalign*
Here I use the recursive formula $$ binomnk = binomn-1k+binomn-1k-1 $$
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
$endgroup$
add a comment
|
$begingroup$
If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$)
$$ sum_n=0^1 binom1n s^n = 1+s=(1+s)^1$$
($N=m to N=m+1$)
beginalign* (1+s)^N+1 &= (1+s)(1+s)^N=(1+s)sum_n=0^N binomNn s^n \&= sum_n=0^N binomNn s^n+sum_n=0^N binomNn s^n+1 \ &= 1+sum_n=1^N binomNn s^n + sum_n=1^N binomNn-1 s^n + s^N+1 \ &= 1+sum_n=1^N left( binomNn + binomNn-1 right) s^n + s^N+1 \ &= 1+sum_n=1^N binomN+1n s^n + s^N+1 \&
= sum_n=0^N+1 binomN+1n s^n
endalign*
Here I use the recursive formula $$ binomnk = binomn-1k+binomn-1k-1 $$
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
$endgroup$
add a comment
|
$begingroup$
If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$)
$$ sum_n=0^1 binom1n s^n = 1+s=(1+s)^1$$
($N=m to N=m+1$)
beginalign* (1+s)^N+1 &= (1+s)(1+s)^N=(1+s)sum_n=0^N binomNn s^n \&= sum_n=0^N binomNn s^n+sum_n=0^N binomNn s^n+1 \ &= 1+sum_n=1^N binomNn s^n + sum_n=1^N binomNn-1 s^n + s^N+1 \ &= 1+sum_n=1^N left( binomNn + binomNn-1 right) s^n + s^N+1 \ &= 1+sum_n=1^N binomN+1n s^n + s^N+1 \&
= sum_n=0^N+1 binomN+1n s^n
endalign*
Here I use the recursive formula $$ binomnk = binomn-1k+binomn-1k-1 $$
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
$endgroup$
If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$)
$$ sum_n=0^1 binom1n s^n = 1+s=(1+s)^1$$
($N=m to N=m+1$)
beginalign* (1+s)^N+1 &= (1+s)(1+s)^N=(1+s)sum_n=0^N binomNn s^n \&= sum_n=0^N binomNn s^n+sum_n=0^N binomNn s^n+1 \ &= 1+sum_n=1^N binomNn s^n + sum_n=1^N binomNn-1 s^n + s^N+1 \ &= 1+sum_n=1^N left( binomNn + binomNn-1 right) s^n + s^N+1 \ &= 1+sum_n=1^N binomN+1n s^n + s^N+1 \&
= sum_n=0^N+1 binomN+1n s^n
endalign*
Here I use the recursive formula $$ binomnk = binomn-1k+binomn-1k-1 $$
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
answered 11 hours ago
serasera
7372 silver badges18 bronze badges
7372 silver badges18 bronze badges
add a comment
|
add a comment
|
$begingroup$
By the Binomial Theorem
$$(a+b)^N=sum_n=0^NNchoose na^ncdot b^N-n$$
choosing $a=s$ and $b=1$ leads to
$$(1+s)^N=(s+1)^N=sum_n=0^NNchoose ns^ncdot 1^N-n=sum_n=0^NNchoose ns^n$$
$endgroup$
add a comment
|
$begingroup$
By the Binomial Theorem
$$(a+b)^N=sum_n=0^NNchoose na^ncdot b^N-n$$
choosing $a=s$ and $b=1$ leads to
$$(1+s)^N=(s+1)^N=sum_n=0^NNchoose ns^ncdot 1^N-n=sum_n=0^NNchoose ns^n$$
$endgroup$
add a comment
|
$begingroup$
By the Binomial Theorem
$$(a+b)^N=sum_n=0^NNchoose na^ncdot b^N-n$$
choosing $a=s$ and $b=1$ leads to
$$(1+s)^N=(s+1)^N=sum_n=0^NNchoose ns^ncdot 1^N-n=sum_n=0^NNchoose ns^n$$
$endgroup$
By the Binomial Theorem
$$(a+b)^N=sum_n=0^NNchoose na^ncdot b^N-n$$
choosing $a=s$ and $b=1$ leads to
$$(1+s)^N=(s+1)^N=sum_n=0^NNchoose ns^ncdot 1^N-n=sum_n=0^NNchoose ns^n$$
answered 11 hours ago
community wiki
b00n heT
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%2f3360862%2fcould-someone-please-show-me-the-steps-of-this-sum%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
$begingroup$
It should be $(1+s)^N$.
$endgroup$
– Wuestenfux
11 hours ago
$begingroup$
It is a finite sum. Convergence makes sense only for infinite sums.
$endgroup$
– Somos
11 hours ago
$begingroup$
Isn't that the definition of "binomial coefficients"?
$endgroup$
– mr_e_man
14 mins ago