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I have tried trying to find a pattern but i don't believe that the right way. If you help me it would be great.

Camilacol
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  • I'm sorry, let me correct it – Camilacol Apr 27 '21 at 00:57
  • Thanks for fixing it. Do you know about "combinatorial proofs"? I.e. counting the same set of objects in two different ways, which establishes a formula? – wormram Apr 27 '21 at 00:59
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    Actually, I do not know any book of the topic, if you have one it would be great. I'm a pioneer in this proofs. – Camilacol Apr 27 '21 at 01:01
  • What I wrote is the only information they gave me. – Camilacol Apr 27 '21 at 01:03
  • This isn't a direct answer to your problem, but the following reference may be useful: http://discrete.openmathbooks.org/dmoi2/sec_comb-proofs.html it shows how to prove identities similar to this one. The technique described there should be helpful in this case. – wormram Apr 27 '21 at 01:04

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Recall that the binomial formula states that $$(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k} $$ Differentiating with respect to $x$ yields $$n(x+y)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kx^{k-1}y^{n-k}$$

Then taking $x=y=1$, the result follows.

C Squared
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