Proving Binomial Sums: Step-by-Step Guide for Solving Homework Equations"

[JP16]

Homework Statement

https://www.physicsforums.com/attachment.php?attachmentid=39642&stc=1&d=1317853920

how do you go about solving this?

Homework Equations

i have proved the binomial theorem.

The Attempt at a Solution

i was considering cases, for j(even or odd). would this be the right direction?

 

[Ray Vickson]

Homework Statement

https://www.physicsforums.com/attachment.php?attachmentid=39642&stc=1&d=1317853920

how do you go about solving this?

Homework Equations

i have proved the binomial theorem.

The Attempt at a Solution

i was considering cases, for j(even or odd). would this be the right direction?

Do you know the binomial expansion of (1+x)^n?

RGV

 

[JP16]

yes, how can i use that to prove it?

 

[JP16]

oh ! (1-1)^n = 0, wow that was really simple. Thank you for the help!

 

[JP16]

how about for this?

https://www.physicsforums.com/attachment.php?attachmentid=39650&stc=1&d=1317861268

(a+b)ⁿ = Ʃ_{l odd} ($^{n}_{l}$) a^lb^n-l where l is odd. where do i go from there?

 

[Dick]

Try subtracting the binomial expansion of (1-1)^n from (1+1)^n. What happens to the even and odd terms?

 

[JP16]

$\sum^{n}_{j=0}$ (-1)^j($^{n}_{j}$) - $\sum^{n}_{j=0}$($^{n}_{j}$)
how do you simplify this? I am sorry, i have really been struggling on this!

edit: oh would this only give you the even terms?

 

[Dick]

$\sum^{n}_{j=0}$ (-1)^j($^{n}_{j}$) - $\sum^{n}_{j=0}$($^{n}_{j}$)
how do you simplify this? I am sorry, i have really been struggling on this!

edit: oh would this only give you the even terms?

No, I think it will only give you the odd terms. Twice. Sort of. Keep thinking. The even terms will cancel.

 

[JP16]

oh yes, i was overlooking the subtraction, and thinking of sum, hence the even terms. But yes they would only give the odd terms. How would it give me it twice though && how do i go about simplifying? Let's say:#of_term...(1+1)^n...(1-1)^n....difference
0......1......0......1

what would be the second case? since (1-1)^n would always be 0?

 

[Dick]

oh yes, i was overlooking the subtraction, and thinking of sum, hence the even terms. But yes they would only give the odd terms. How would it give me it twice though && how do i go about simplifying? Let's say:#of_term...(1+1)^n...(1-1)^n....difference
0......1......0......1

what would be the second case? since (1-1)^n would always be 0?

Not sure what you are getting at there. (-1)^j*C(n,j)-C(n,j)=(-2)*C(n,j) if j is odd. Otherwise it's zero. That gives you a sum over all odd binomial coefficients doesn't it? Now you just have to figure out what it's equal to. What are (1+1)^n and (1-1)^n? Don't forget where this expression is coming from.

 

[JP16]

Not sure what you are getting at there. (-1)^j*C(n,j)-C(n,j)=(-2)*C(n,j) if j is odd. Otherwise it's zero. That gives you a sum over all odd binomial coefficients doesn't it? Now you just have to figure out what it's equal to. What are (1+1)^n and (1-1)^n? Don't forget where this expression is coming from.

how did you get the -2?

(1+1)^n = C(n,j) = 2^n

(1-1)^n = (-1)^j*C(n,j) = 0

ah! this is really confusing.

 

[Dick]

how did you get the -2?

(1+1)^n = C(n,j) = 2^n

(1-1)^n = (-1)^j*C(n,j) = 0

ah! this is really confusing.

(-1)^j*C(n,j)-C(n,j) is 0 if j is even, if j is odd (-1)^j*C(n,j)-C(n,j)=(-2)*C(n,j). Isn't it? I'm not sure what's confusing about that.

 

[JP16]

ohh then
(-2)*C(n,j) = (-2)*(1+1)^n ?

 

[Dick]

ohh then
(-2)*C(n,j) = (-2)*(1+1)^n ?

Ack. Where did you get the (-2) on the right side? And you mean sum over j odd on the left side, right? Go back and think about this again. If you make me comment about this again, you might not like the comment. So don't. Think hard first. I know it's late, but I think you can do it.

 

[JP16]

oh it is soo much clear after writing it in factorial way. so:

$\frac{(-1)^jn!}{j!(n-j)!}$ - $\frac{n!}{j!(n-j)!}$ = $\frac{n!(-1^j - 1)}{j!(n-j)!}$

where [(-1)^j -1] = -2 since j is odd.

and so yes, then it is (-2)*C(n,j)

 

[Dick]

oh it is soo much clear after writing it in factorial way. so:

$\frac{(-1)^jn!}{j!(n-j)!}$ - $\frac{n!}{j!(n-j)!}$ = $\frac{n!(-1^j - 1)}{j!(n-j)!}$

where [(-1)^j -1] = -2 since j is odd.

Ok, so you've got it then, right? Sure (-1)^j*a-a=(-2)*a if j is odd and 0 if j is even. Doesn't matter much what 'a' is, but if putting the factorials in makes you happy, it's fine with me.

 

[JP16]

i kind of realize that now, but i just couldn't see it with the summations.

What are (1+1)^n and (1-1)^n?

no, i still don't know where to go afterwards. what do you mean by "what are"?

sorry, i am just as much frustrated.

 

[Dick]

i kind of realize that now, but i just couldn't see it with the summations.



no, i still don't know where to go afterwards. what do you mean by "what are"?

sorry, i am just as much frustrated.

By "what are" I'm looking for an answer like (1-1)^n=0. You were trying to express (1-1)^n-(1+1)^n in terms of binomial coefficients in hopes of answering your question about the sum of the odd coefficients. Remember?