Hi. My teacher told me I should try getting help from this forum with a proof I'm stuck with and here I am. I need this proof for my personal research (I'm not in a university) in mathematics.

So I am having a hard time proving this equation: puu.sh/2VUM9.png (sorry for the url, I'm too lazy to type the equation without LaTeX).

I have tried proving the equation by induction but I am stuck at one point.

Here is what I have done so far: puu.sh/2VVfc.jpg (I know it is another link but there is no way I am going to type that here). I might not have used the correct or most commons English mathematical terms and/or words in my work but I think you'll get what I've done so far.

As you may see from the imahe I am stuck with the last part in the proof of inductive statement, which I was hoping someone would finish for me.

If any of you use LaTeX I have uploaded the document which contains the unfinished proof in several formats (for example *.tex) here: puu.sh/2VVAd.zip

It should also be noted that your proof doesn't have to be done by induction. My teacher just suggested it to me and I gave it my best shot.

Also, if you prefer you can prove this equation ( Nexus Number, puu.sh/2VVIj.png ) which is the same equation as mine, but where n-3=d, c=k, i-1=n (is a variable not the imaginary unit).

Lastly why do I think there even is a proof for the equation? That is because Wolframalpha says it is true and even has an article about this equation (Nexus Number, google it), without a proof though.

## Help needed for a mathematical proof

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- Voekoevaka
**Posts:**42**Joined:**Wed Apr 10, 2013 10:29 am UTC**Location:**Over nine thousand.

### Re: Help needed for a mathematical proof

I havn't time to try it, but maybe the Binomial Theorem can help you.

I'm a dozenalist and a believer in Tau !

- Lopsidation
**Posts:**183**Joined:**Tue Oct 27, 2009 11:29 pm UTC

### Re: Help needed for a mathematical proof

Currently, where you are in your inductive proof, you've effectively split the sum you want into two sums. Look at the second sum - it's the same as the one in your inductive hypothesis, just with an extra term for c=k-2, right? You can handle that.

The first sum is a little trickier, but it falls to the same idea. (HINT: what happens when you divide by (i-1)?)

As Voekoevaka says, if you know the Binomial Theorem, you can prove your Nexus Number equation without induction, by writing down the expansion for (n+1)

The first sum is a little trickier, but it falls to the same idea. (HINT: what happens when you divide by (i-1)?)

As Voekoevaka says, if you know the Binomial Theorem, you can prove your Nexus Number equation without induction, by writing down the expansion for (n+1)

^{(d+1)}.-
**Posts:**563**Joined:**Tue Jul 27, 2010 8:48 am UTC

### Re: Help needed for a mathematical proof

Replace n-2 with n, and i-1 with i, and you end up with (i+1)

^{n}-i^{n}=sum(1<=i<=n-1,(n c)i^{c}), adding i^{n}to both sides gives: (i+1)^{n}=sum(1<=i<=n,(n c)i^{c}), which is the regular binomial formula.I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

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