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I'm solving some exercises from Kreszig's Advanced Math book and I got stuck in one:

(10th ed, chapter 15.3, problem 18): Using $(1+z)^p*(1+z)^q=(1+z)^{p+q}$, obtain the basic relation: $$\sum_{n=0}^r \binom{p}{n}\binom{q}{r-n} = \binom{p+q}{r}$$

The Power Series aspect of the problem is that we must explicitly use the following uniqueness theorem for complex power series:

Let the power series $a_0+a_1z+a_2z^2+...$ and $b_0+b_1z+b_2z^2+...$ both be convergent for |z|$\lt$R where R is positive, and let them both have the same sum for all these z. Then the series are identical, that is, $a_0=b_0, a_1=b_1, a_2=b_2, ^{...}.$

I thought about using the factorial formula of binomial coefficient and also the multiplicative form of it, but I couldn't even get started. I'm not the best at power series, so I'm having trouble where can I find or develop the second sum so I can use the theorem.

If it is of any help, I started developing some terms of the sum to see if I could form some kind of pattern or to use some property, but I also failed.

Thanks in advance.

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Since $(1+z)^p\times(1+z)^q=(1+z)^{p+q}$ and both sides can be expressed as convergent power series for $|x|<1$, we know that both power series have the same coefficents by the quoted result in your question. In particular look at the coefficent of $x^r$ in the expansion of $(1+z)^{p+q}$. This is $\binom{p+q}{r}$. Now look at the coefficent of $x^r$ in the expansion of $(1+z)^p\times(1+z)^q=(1+z)^{p+q}$. Remembering how series are multiplied (cauchy product) we get that it is $$\sum_{n=0}^r \binom{p}{n}\binom{q}{r-n}.$$ The result follows.

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