# Proving Binomial Sums

1. Oct 5, 2011

### JP16

1. The problem statement, all variables and given/known data

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

how do you go about solving this?
2. Relevant equations
i have proved the binomial theorem.

3. The attempt at a solution
i was considering cases, for j(even or odd). would this be the right direction?

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2. Oct 5, 2011

### Ray Vickson

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

RGV

3. Oct 5, 2011

### JP16

yes, how can i use that to prove it?

4. Oct 5, 2011

### JP16

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

5. Oct 5, 2011

### JP16

how about for this?

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

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

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6. Oct 5, 2011

### Dick

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

7. Oct 5, 2011

### JP16

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

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

8. Oct 5, 2011

### Dick

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

9. Oct 6, 2011

### 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? Lets 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?

10. Oct 6, 2011

### Dick

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.

11. Oct 6, 2011

### JP16

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.

12. Oct 6, 2011

### Dick

(-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.

13. Oct 6, 2011

### JP16

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

14. Oct 6, 2011

### Dick

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.

Last edited: Oct 6, 2011
15. Oct 6, 2011

### 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)

16. Oct 6, 2011

### Dick

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.

17. Oct 6, 2011

### JP16

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.

18. Oct 6, 2011

### Dick

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?