Exercise 1.4.19* (Conundrum 1/8)

Question

Show that if $m$ and $n$ are non-negative integers then
$$\sum_{k=0}^n (-1)^k \binom{m+1}{k} \binom{m+n-k}{m}  = 
\left\{
\begin{array}{ll}
1 & 	ext{if } n= 0,\
0 & 	ext{if } n>0.
\end{array}
ight.
$$

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} The first of the more challenging eight final conundrums, without any hint. \begin{proof} By Exercise 7, for all $z \in \C$ such that $|z|<1$, $$\frac{1}{(1-z)^{m+1}} = \sum_{l=0}^\infty \binom{m+l}{l} z^l = \sum_{l=0}^\infty \binom{m+l}{m} z^l .$$ Then \begin{align*} 1 &= (1-z)^{m+1} \frac{1}{(1-z)^{m+1}}\ &=\left[\sum_{k=0}^{m+1} (-1)^k \binom{m+1}{k}z^k ight] \left[\sum_{l=0}^\infty \binom{m+l}{l} z^l ight]\ &=\left[\sum_{k=0}^{\infty} (-1)^k \binom{m+1}{k}z^k ight] \left[\sum_{l=0}^\infty \binom{m+l}{l} z^l ight]\qquad ( ext{since } \binom{m+1}{k} = 0 ext{ if } k>m+1)\ &= \sum_{n = 0}^\infty \left [\sum_{k+l = n} (-1)^k \binom{m+1}{k} \binom{m+l}{l} ight] z^n\ &= \sum_{n = 0}^\infty \left [\sum_{k=0}^n (-1)^k \binom{m+1}{k} \binom{m+n-k}{l} ight] z^n\ \end{align*} Thus, for all $z \in \C$ such that $|z|<1$, $$ 1 = \sum_{n = 0}^\infty \left [\sum_{k=0}^n (-1)^k \binom{m+1}{k} \binom{m+n-k}{l} ight] z^n\ $$ The unicity of the coefficient of $z^n$ in the powers series shows that $$\sum_{k=0}^n (-1)^k \binom{m+1}{k} \binom{m+n-k}{m} = \left\{ \begin{array}{ll} 1 & ext{if } n= 0,\ 0 & ext{if } n>0. \end{array} ight. $$ \end{proof}

Exercise 1.4.19* (Conundrum 1/8)

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Exercise 1.4.19* (Conundrum 1/8)

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