# Binomial Theorem proof by induction, Spivak

1. Jul 7, 2012

### usn7564

The problem statement, all variables and given/known data
Prove the binomial theorem by induction.

The attempt at a solution
http://desmond.imageshack.us/Himg35/scaled.php?server=35&filename=sumu.png&res=landing [Broken]

Hi, running into trouble with this proof and google hasn't helped me. I don't understand the jump here, and as it's not really explained in the videos/.pdf's I found I presume it's something really simple that I'm just not getting.

I saw one explanation saying "just put k=j+1, put k in the equation in place of j then switch back to j" and while it worked I'm not seeing why I can replace j with k, as k=/= j.

Any input would be appreciated, this is really puzzling me.

Last edited by a moderator: May 6, 2017
2. Jul 7, 2012

### Curious3141

j and k are just dummy variables in the sum. Once you've replaced j with k, and worked out the form of the summand and the bounds, you can replace k with anything you like, including j. It's OK as long as you replace every instance of it.

Try replacing the "k"s with the actual numbers and write out the sum. Do the same with the "j"s and you'll find out the come out to be the sum of the same terms, proving they're equal.

Last edited by a moderator: May 6, 2017
3. Jul 7, 2012

### usn7564

Ah, right. Thought I wouldn't be able to without doing it on LHS as well, suppose it shouldn't make a difference though. I'll try putting some numbers in and hopefully have it 100% clear. Thanks!

4. Jul 7, 2012

### awkward

I think you will need to use the following identity:
$$\binom{n}{j-1} + \binom{n}{j} = \binom{n+1}{j}$$

It shouldn't be hard to prove if you haven't seen it before.

5. Jul 7, 2012

6. Jul 8, 2012

### SammyS

Staff Emeritus
This may not be anymore helpful to you than the above posts, but I'll add my two cents.

How is $\displaystyle \sum_{j=0}^{n} \binom{n}{j}\,a^{(n-j)}\,b^{\,(j+1)} \$ equal to $\displaystyle \sum_{j=1}^{n+1} \binom{n}{j-1}\,a^{(n+1-j)}\,b^{\,j} \ ?$​
Starting with $\displaystyle \sum_{j=0}^{n} \binom{n}{j}\,a^{(n-j)}\,b^{\,(j+1)} \,,$ let k = j+1. Then j = k-1 . The sum starts with j = 0, which corresponds to k = 1 . The sum terminates with j = n, which corresponds to k = n+1 .
Replacing j with k-1 gives: $\displaystyle \sum_{k=1}^{n+1} \binom{n}{k-1}\,a^{(n-(k-1))}\,b^{\,(k-1+1)} \,.$
Simplifying this, we have: $\displaystyle \sum_{k=1}^{n+1} \binom{n}{k-1}\,a^{(n+1-k)}\,b^{\,k} \,.$