<s>[INST] How to prove this combinatorially $\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+\cdots+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$? $n,m,k$ are natural numbers.
$$\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+\cdots+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$$
I need to prove this combinatorially but I can't think of a story, how can I approach this? I thought about starting from the left side
 [/INST] ${n \choose r} + {n \choose r+1} = {n+1 \choose r+1}$

${n \choose k} = {n+1 \choose k+1} - {n \choose k+1}$ .....(1)
${n+1 \choose k} = {n+2 \choose k+1} - {n+2 \choose k+1}$.....(2)
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${n+m \choose k} = {n+m+1 \choose k+1} - {n+m\choose k}$.....(m)

(1) + (2).............(m)  implies
$\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+...+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$
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