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Hockey Stick Identity.md

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+#Math #Probability
+
+# Statement
+
+For $n \geq r$, $n, r \in \mathbb{N}$:
+
+$$
+\sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1}
+$$
+
+# Proof
+
+Let us have a base case $n = r$:
+
+$$
+{r \choose r} = {r + 1 \choose r + 1} = 1
+$$
+
+Now suppose $\sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1}$ for a certain $n$:
+
+$$
+\sum _{i = r}^n {i \choose r} + {n + 1 \choose r} \\
+= {n + 1 \choose r + 1} + {n + 1 \choose r} \\
+= {n + 2 \choose r + 1}
+$$
+
+Since $n = r$ is true, and if a case is true for $n$, it is true for $n + 1$, this statement is true for all $n \geq r$.